print binary tree level by level in python

Question

I want to print my binary tree in the following manner:

 10
 6 12
 5 7 11 13 

I have written code for insertion of nodes but can't able to write for printing the tree. so please help on this . My code is :

class Node:
 def __init__(self,data):
 self.data=data
 self.left=None
 self.right=None
 self.parent=None
class binarytree:
 def __init__(self):
 self.root=None
 self.size=0
 def insert(self,data):
 if self.root==None:
 self.root=Node(data)
 else:
 current=self.root
 while 1:
 if data < current.data:
 if current.left:
 current=current.left
 else:
 new=Node(data)
 current.left=new
 break;
 elif data > current.data:
 if current.right:
 current=current.right
 else:
 new=Node(data)
 current.right=new
 break;
 else:
 break
 b=binarytree() 

• user5545202: You could add a method to your Node class that counts counts how many times you can return a parent before you get None or your root.
• AKS: Just FYSA In BASH, I used a .txt file approach and tried for printing the same from any root node here: stackoverflow.com/a/67341334/1499296

Answer 1

Here's my attempt, using recursion, and keeping track of the size of each node and the size of children.

class BstNode:
 def __init__(self, key):
 self.key = key
 self.right = None
 self.left = None
 def insert(self, key):
 if self.key == key:
 return
 elif self.key < key:
 if self.right is None:
 self.right = BstNode(key)
 else:
 self.right.insert(key)
 else: # self.key > key
 if self.left is None:
 self.left = BstNode(key)
 else:
 self.left.insert(key)
 def display(self):
 lines, *_ = self._display_aux()
 for line in lines:
 print(line)
 def _display_aux(self):
 """Returns list of strings, width, height, and horizontal coordinate of the root."""
 # No child.
 if self.right is None and self.left is None:
 line = '%s' % self.key
 width = len(line)
 height = 1
 middle = width // 2
 return [line], width, height, middle
 # Only left child.
 if self.right is None:
 lines, n, p, x = self.left._display_aux()
 s = '%s' % self.key
 u = len(s)
 first_line = (x + 1) * ' ' + (n - x - 1) * '_' + s
 second_line = x * ' ' + '/' + (n - x - 1 + u) * ' '
 shifted_lines = [line + u * ' ' for line in lines]
 return [first_line, second_line] + shifted_lines, n + u, p + 2, n + u // 2
 # Only right child.
 if self.left is None:
 lines, n, p, x = self.right._display_aux()
 s = '%s' % self.key
 u = len(s)
 first_line = s + x * '_' + (n - x) * ' '
 second_line = (u + x) * ' ' + '\\' + (n - x - 1) * ' '
 shifted_lines = [u * ' ' + line for line in lines]
 return [first_line, second_line] + shifted_lines, n + u, p + 2, u // 2
 # Two children.
 left, n, p, x = self.left._display_aux()
 right, m, q, y = self.right._display_aux()
 s = '%s' % self.key
 u = len(s)
 first_line = (x + 1) * ' ' + (n - x - 1) * '_' + s + y * '_' + (m - y) * ' '
 second_line = x * ' ' + '/' + (n - x - 1 + u + y) * ' ' + '\\' + (m - y - 1) * ' '
 if p < q:
 left += [n * ' '] * (q - p)
 elif q < p:
 right += [m * ' '] * (p - q)
 zipped_lines = zip(left, right)
 lines = [first_line, second_line] + [a + u * ' ' + b for a, b in zipped_lines]
 return lines, n + m + u, max(p, q) + 2, n + u // 2
import random
b = BstNode(50)
for _ in range(50):
 b.insert(random.randint(0, 100))
b.display()

Example output:

 __50_________________________________________ 
 / \
 ________________________43_ ________________________99
 / \ / 
 _9_ 48 ____________67_____________________ 
 / \ / \ 
 3 11_________ 54___ ______96_ 
/ \ \ \ / \ 
0 8 ____26___________ 61___ ________88___ 97 
 / \ / \ / \ 
 14_ __42 56 64_ 75_____ 92_ 
 / \ / / \ / \ / \ 
 13 16_ 33_ 63 65_ 72 81_ 90 94 
 \ / \ \ / \ 
 25 __31 41 66 80 87 
 / / 
 28_ 76 
 \ 
 29 

• Apostolos: However, it can't display an already created BT, can it? Even using the 'insert()' function, new nodes are inserted according to their value not as one would like to.
• J. V.: @Apostolos not sure I understand your comment. How else would one like to insert nodes?
• Apostolos: Example: a BT set as follows: tree = (50, (30, (20, (10, None, None), None), (40, None, None)), (80, (70, (60, None, (61, None, (62, None, None))), None), (90, None, None))). It is fully defined, i.e. all left and right positions of the children's are set (even if empty). Does this setting make sense to you?
• J. V.: The tree structure depends on the order in which you added elements, but in your case if you add them the way you listed them you get what you want: b = BstNode(50); for n in (30, 20, 10, 40, 80, 70, 60, 61, 62, 90): b.insert(n); b.display()
• J. V.: Of course it doesn't show None but you can modify _display_aux to show it if that's what you want

Answer 2

class Node(object):
 def __init__(self, value, left=None, right=None):
 self.value = value
 self.left = left
 self.right = right
 
def printTree(node, level=0):
 if node != None:
 printTree(node.left, level + 1)
 print(' ' * 4 * level + '-> ' + str(node.value))
 printTree(node.right, level + 1)
t = Node(1, Node(2, Node(4, Node(7)),Node(9)), Node(3, Node(5), Node(6)))
printTree(t)

output:

 -> 7
 -> 4
 -> 2
 -> 9
-> 1
 -> 5
 -> 3
 -> 6

• AeroHil: After switching up the left and right nodes from the recursion calls, I was able to get a better representation of the tree (after a rotation of 90 degree clockwise). This was the best quick-and-dirty way for me to print and display a tree.
• Eratosthenes: Not bad, but this output incorrectly prints the mirror image of the tree.

Answer 3

What you're looking for is breadth-first traversal, which lets you traverse a tree level by level. Basically, you use a queue to keep track of the nodes you need to visit, adding children to the back of the queue as you go (as opposed to adding them to the front of a stack). Get that working first.

After you do that, then you can figure out how many levels the tree has (log2(node_count) + 1) and use that to estimate whitespace. If you want to get the whitespace exactly right, you can use other data structures to keep track of how many spaces you need per level. A smart estimation using number of nodes and levels should be enough, though.

Answer 4

I am leaving here a stand-alone version of @J. V.'s code. If anyone wants to grab his/her own binary tree and pretty print it, pass the root node and you are good to go.

If necessary, change val, left and right parameters according to your node definition.

def print_tree(root, val="val", left="left", right="right"):
 def display(root, val=val, left=left, right=right):
 """Returns list of strings, width, height, and horizontal coordinate of the root."""
 # No child.
 if getattr(root, right) is None and getattr(root, left) is None:
 line = '%s' % getattr(root, val)
 width = len(line)
 height = 1
 middle = width // 2
 return [line], width, height, middle
 # Only left child.
 if getattr(root, right) is None:
 lines, n, p, x = display(getattr(root, left))
 s = '%s' % getattr(root, val)
 u = len(s)
 first_line = (x + 1) * ' ' + (n - x - 1) * '_' + s
 second_line = x * ' ' + '/' + (n - x - 1 + u) * ' '
 shifted_lines = [line + u * ' ' for line in lines]
 return [first_line, second_line] + shifted_lines, n + u, p + 2, n + u // 2
 # Only right child.
 if getattr(root, left) is None:
 lines, n, p, x = display(getattr(root, right))
 s = '%s' % getattr(root, val)
 u = len(s)
 first_line = s + x * '_' + (n - x) * ' '
 second_line = (u + x) * ' ' + '\\' + (n - x - 1) * ' '
 shifted_lines = [u * ' ' + line for line in lines]
 return [first_line, second_line] + shifted_lines, n + u, p + 2, u // 2
 # Two children.
 left, n, p, x = display(getattr(root, left))
 right, m, q, y = display(getattr(root, right))
 s = '%s' % getattr(root, val)
 u = len(s)
 first_line = (x + 1) * ' ' + (n - x - 1) * '_' + s + y * '_' + (m - y) * ' '
 second_line = x * ' ' + '/' + (n - x - 1 + u + y) * ' ' + '\\' + (m - y - 1) * ' '
 if p < q:
 left += [n * ' '] * (q - p)
 elif q < p:
 right += [m * ' '] * (p - q)
 zipped_lines = zip(left, right)
 lines = [first_line, second_line] + [a + u * ' ' + b for a, b in zipped_lines]
 return lines, n + m + u, max(p, q) + 2, n + u // 2
 lines, *_ = display(root, val, left, right)
 for line in lines:
 print(line)

---

print_tree(root)
 __7 
 / \
 ___10_ 3
 / \ 
 _19 13 
 / \ 
 9 8_ 
/ \ \ 
4 0 12 

• NoPlaceLike127.0.0.1: nice work! you've got a bug in your display() args however. Should look like this: python def print_tree(root, val="val", left="left", right="right"): def display(root, val=val, left=left, right=right):
• squall: @BcK hi, I've just found your printing def, as I also have to implement this kind of printing, but I have a problem with changing parameters - you said we'd take care of left and right but I don't know what to change it for. Could you check that in your free time?: stackoverflow.com/questions/67530220/…

Answer 5

Simple solution with no recursion

def PrintTree(root):
 def height(root):
 return 1 + max(height(root.left), height(root.right)) if root else -1 
 nlevels = height(root)
 width = pow(2,nlevels+1)
 q=[(root,0,width,'c')]
 levels=[]
 while(q):
 node,level,x,align= q.pop(0)
 if node: 
 if len(levels)<=level:
 levels.append([])
 
 levels[level].append([node,level,x,align])
 seg= width//(pow(2,level+1))
 q.append((node.left,level+1,x-seg,'l'))
 q.append((node.right,level+1,x+seg,'r'))
 for i,l in enumerate(levels):
 pre=0
 preline=0
 linestr=''
 pstr=''
 seg= width//(pow(2,i+1))
 for n in l:
 valstr= str(n[0].val)
 if n[3]=='r':
 linestr+=' '*(n[2]-preline-1-seg-seg//2)+ ' ̄'*(seg +seg//2)+'\\'
 preline = n[2] 
 if n[3]=='l':
 linestr+=' '*(n[2]-preline-1)+'/' + ' ̄'*(seg+seg//2) 
 preline = n[2] + seg + seg//2
 pstr+=' '*(n[2]-pre-len(valstr))+valstr #correct the potition acording to the number size
 pre = n[2]
 print(linestr)
 print(pstr) 

Sample output

 1
 / ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄\
 2 3
 / ̄ ̄ ̄ ̄ ̄ ̄\ / ̄ ̄ ̄ ̄ ̄ ̄\
 4 5 6 7
 / ̄ ̄\ / ̄ / ̄
 8 9 10 12

• Yuri Popiv: masterpiece, thanks

Answer 6

I enhanced Prashant Shukla answer to print the nodes on the same level in the same line without spaces.

class Node(object):
 def __init__(self, value, left=None, right=None):
 self.value = value
 self.left = left
 self.right = right
 def __str__(self):
 return str(self.value)
def traverse(root):
 current_level = [root]
 while current_level:
 print(' '.join(str(node) for node in current_level))
 next_level = list()
 for n in current_level:
 if n.left:
 next_level.append(n.left)
 if n.right:
 next_level.append(n.right)
 current_level = next_level
t = Node(1, Node(2, Node(4, Node(7)), Node(9)), Node(3, Node(5), Node(6)))
traverse(t)

• Vladislavs Dovgalecs: I am confused about current_level that is overwritten in the for loop over the elements of current_level! Can you explain how this piece of code works?

Answer 7

Just use this small method of print2DTree:

class bst:
 def __init__(self, value):
 self.value = value
 self.right = None
 self.left = None
 
def insert(root, key):
 if not root:
 return bst(key)
 if key >= root.value:
 root.right = insert(root.right, key)
 elif key < root.value:
 root.left = insert(root.left, key)
 return root
def insert_values(root, values):
 for value in values:
 root = insert(root, value)
 return root
def print2DTree(root, space=0, LEVEL_SPACE = 5):
 if (root == None): return
 space += LEVEL_SPACE
 print2DTree(root.right, space)
 # print() # neighbor space
 for i in range(LEVEL_SPACE, space): print(end = " ") 
 print("|" + str(root.value) + "|<")
 print2DTree(root.left, space)
root = insert_values(None, [8, 4, 12, 2, 6, 10, 14, 1, 3, 5, 7, 9, 11, 13, 15])
print2DTree(root) 

Results:

Example Tree

Example 2D print of Tree

Answer 8

Here's a 2-pass solution with no recursion for general binary trees where each node has a value that "fits" within the allotted space (values closer to the root have more room to spare). (Pass 0 computes the tree height).

'''
0: 0
1: 1 2
2: 3 4 5 6
3: 7 8 9 a b c d e
h: 4
N: 2**4 - 1 <--| 2**0 + 2**1 + 2**2 + 2**3
'''
import math
def t2_lvl( i): return int(math.log2(i+1)) if 0<i else 0 # @meta map the global idx to the lvl
def t2_i2base(i): return (1<<t2_lvl(i))-1 # @meta map the global idx to the local idx (ie. the idx of elem 0 in the lvl at idx @i)
def t2_l2base(l): return (1<< l) -1 # @meta map the lvl to the local idx (ie. the idx of elem 0 in lvl @l)
class Tree2: # @meta a 2-tree is a tree with at most 2 sons per dad
 def __init__(self, v=None):
 self.v = v
 self.l = None
 self.r = None
 def __str__(self): return f'{self.v}'
def t2_show(tree:Tree2): # @meta 2-pass fn. in the 1st pass we compute the height
 if not tree: return
 q0 = [] # perm queue
 q1 = [] # temp queue
 # pass 0
 h = 0 # height is the number of lvls
 q0.append((tree,0))
 q1.append((tree,0))
 while q1:
 n,i = q1.pop(0)
 h = max(h, t2_lvl(i))
 if n.l: l=(n.l, 2*i+1); q0.append(l); q1.append(l)
 if n.r: r=(n.r, 2*i+2); q0.append(r); q1.append(r)
 h += 1 # nlvls
 N = 2**h - 1 # nelems (for a perfect tree of this height)
 W = 1 # elem width
 # pass 1
 print(f'\n\x1b[31m{h} \x1b[32m{len(q0)}\x1b[0m')
 print(f'{0:1x}\x1b[91m:\x1b[0m',end='')
 for idx,(n,i) in enumerate(q0):
 l = t2_lvl(i) # lvl
 b = (1<<l)-1 # base
 s0 = (N // (2**(l+1)))
 s1 = (N // (2**(l+0)))
 s = 3+1 + s0 + (i-b)*(s1+1) # absolute 1-based position (from the beginning of line)
 w = int(2**(h-l-2)) # width (around the element) (to draw the surrounding @-)
 # print(f'{i:2x} {l} {i-b} {s0:2x} {s1:2x} {s:2x} {w:x} {n.v:02x}')
 if 0<idx and t2_lvl(q0[idx-1][1])!=l: print(f'\n{l:1x}\x1b[91m:\x1b[0m',end='') # new level: go to the next line
 print(f"\x1b[{s-w}G{w*'-'}\x1b[1G", end='')
 print(f"\x1b[{s}G{n.v:1x}\x1b[1G", end='') # `\x1b[XG` is an ANSI escape code that moves the cursor to column X
 print(f"\x1b[{s+W}G{w*'-'}\x1b[1G", end='')
 print()

And an example:

tree = Tree2(0)
tree.l = Tree2(1)
tree.r = Tree2(2)
tree.l.l = Tree2(3)
tree.r.l = Tree2(4)
tree.r.r = Tree2(5)
tree.l.l.l = Tree2(3)
tree.r.l.l = Tree2(6)
tree.r.l.r = Tree2(7)
tree.l.l.l.l = Tree2(3)
tree.r.l.l.l = Tree2(8)
tree.r.l.l.r = Tree2(9)
t2_show(tree)

Output:

5 12
0: --------0--------
1: ----1---- ----2----
2: --3-- --4-- --5--
3: -3- -6- -7-
4: 3 8 9

Another output example:

7 127
0: --------------------------------0--------------------------------
1: ----------------1---------------- ----------------2----------------
2: --------3-------- --------4-------- --------5-------- --------6--------
3: ----7---- ----8---- ----9---- ----a---- ----b---- ----c---- ----d---- ----e----
4: --f-- --0-- --1-- --2-- --3-- --4-- --5-- --6-- --7-- --8-- --9-- --a-- --b-- --c-- --d-- --e--
5: -f- -0- -1- -2- -3- -4- -5- -6- -7- -8- -9- -a- -b- -c- -d- -e- -f- -0- -1- -2- -3- -4- -5- -6- -7- -8- -9- -a- -b- -c- -d- -e-
6: f 0 1 2 3 4 5 6 7 8 9 a b c d e f 0 1 2 3 4 5 6 7 8 9 a b c d e f 0 1 2 3 4 5 6 7 8 9 a b c d e f 0 1 2 3 4 5 6 7 8 9 a b c d e

Answer 9

code Explanation:

• by using the BFS get the lists of list contains elements of each level
• number of white spaces at any level = (max number of element in tree)//2^level
• maximum number of elements of h height tree = 2^h -1; considering root level height as 1
• print the value and white spaces find my Riple.it link here print-bst-tree

def bfs(node,level=0,res=[]):
 if level<len(res):
 if node:
 res[level].append(node.value)
 else:
 res[level].append(" ")
 else:
 if node:
 res.append([node.value])
 else:
 res.append([" "])
 if not node:
 return 
 bfs(node.left,level+1,res)
 bfs(node.right,level+1,res)
 return res
 
def printTree(node):
 treeArray = bfs(node)
 h = len(treeArray)
 whiteSpaces = (2**h)-1
 
 def printSpaces(n):
 for i in range(n):
 print(" ",end="")
 
 for level in treeArray:
 whiteSpaces = whiteSpaces//2
 for i,x in enumerate(level):
 if i==0:
 printSpaces(whiteSpaces)
 print(x,end="")
 printSpaces(1+2*whiteSpaces)
 print()
#driver Code
printTree(root)

#output Output

Answer 10

I have added this method in my python Node class to be able to print it in any level of the tree

 def print_tree(self, indent=0, is_left=None, prefix=' ', has_right_child=True, has_left_child=True):
 if self.left:
 self.left.print_tree(indent + 1, True, prefix + (' ' if has_right_child else '| '), has_right_child=True, has_left_child=False)
 if is_left == True:
 if self.left:
 print(prefix + f'd={self.depth}' + ' | ')
 else:
 print(prefix + f'd={self.depth}' )
 print(prefix + '┌──>' + str(self.value))
 elif is_left == False:
 print(prefix + '└──>' + str(self.value))
 if self.right:
 print(prefix + f'd={self.depth}' + ' | ')
 else:
 print(prefix + f'd={self.depth}' )
 else:
 print('|── ' + str(self.value))
 if self.right:
 self.right.print_tree(indent + 1, False, prefix + (' ' if has_left_child else '| '), has_right_child=False, has_left_child=True)

My Node class

class Node:
 # Classic Node class with left, right and value, depth is Optional
 def __init__(self, value=None, left=None, right=None, depth=None):
 self.value = value
 self.left = left
 self.right = right
 self.depth = depth

Example Output

 d=1
 ┌──>1 
 |── 0
 | d=2
 | ┌──>156
 └──>23
 d=1 | 
 | d=3
 | ┌──>213
 └──>321
 d=2 | 
 | d=4
 | ┌──>245
 └──>123
 d=3 | 
 └──>123
 d=4

d means depth

Answer 11

class magictree:
 def __init__(self, parent=None):
 self.parent = parent
 self.level = 0 if parent is None else parent.level + 1
 self.attr = []
 self.rows = []
 def add(self, value):
 tr = magictree(self)
 tr.attr.append(value)
 self.rows.append(tr)
 return tr
 def printtree(self):
 def printrows(rows):
 for i in rows:
 print("{}{}".format(i.level * "\t", i.attr))
 printrows(i.rows)
 printrows(self.rows)
tree = magictree()
group = tree.add("company_1")
group.add("emp_1")
group.add("emp_2")
emp_3 = group.add("emp_3")
group = tree.add("company_2")
group.add("emp_5")
group.add("emp_6")
group.add("emp_7")
emp_3.add("pencil")
emp_3.add("pan")
emp_3.add("scotch")
tree.printtree()

---

result:

['company_1']
 ['emp_1']
 ['emp_2']
 ['emp_3']
 ['pencil']
 ['pan']
 ['scotch']
['company_2']
 ['emp_5']
 ['emp_6']
 ['emp_7']

Answer 12

As I came to this question from Google (and I bet many others did too), here is binary tree that has multiple children, with a print function (__str__ which is called when doing str(object_var) and print(object_var)).

Code:

from typing import Union, Any
class Node:
 def __init__(self, data: Any):
 self.data: Any = data
 self.children: list = []
 
 def insert(self, data: Any):
 self.children.append(Node(data))
 def __str__(self, top: bool=True) -> str:
 lines: list = []
 lines.append(str(self.data))
 for child in self.children:
 for index, data in enumerate(child.__str__(top=False).split("\n")):
 data = str(data)
 space_after_line = " " * index
 if len(lines)-1 > index:
 lines[index+1] += " " + data
 if top:
 lines[index+1] += space_after_line
 else:
 if top:
 lines.append(data + space_after_line)
 else:
 lines.append(data)
 for line_number in range(1, len(lines) - 1):
 if len(lines[line_number + 1]) > len(lines[line_number]):
 lines[line_number] += " " * (len(lines[line_number + 1]) - len(lines[line_number]))
 lines[0] = " " * int((len(max(lines, key=len)) - len(str(self.data))) / 2) + lines[0]
 return '\n'.join(lines)
 def hasChildren(self) -> bool:
 return bool(self.children)
 def __getitem__(self, pos: Union[int, slice]):
 return self.children[pos]

And then a demo:

# Demo
root = Node("Languages Good For")
root.insert("Serverside Web Development")
root.insert("Clientside Web Development")
root.insert("For Speed")
root.insert("Game Development")
root[0].insert("Python")
root[0].insert("NodeJS")
root[0].insert("Ruby")
root[0].insert("PHP")
root[1].insert("CSS + HTML + Javascript")
root[1].insert("Typescript")
root[1].insert("SASS")
root[2].insert("C")
root[2].insert("C++")
root[2].insert("Java")
root[2].insert("C#")
root[3].insert("C#")
root[3].insert("C++")
root[0][0].insert("Flask")
root[0][0].insert("Django")
root[0][1].insert("Express")
root[0][2].insert("Ruby on Rails")
root[0][0][0].insert(1.1)
root[0][0][0].insert(2.1)
print(root)

Answer 13

This is part of my own implementation of BST. The ugly part of this problem is that you have to know the space that your children occupies before you can print out yourself. Because you can have very big numbers like 217348746327642386478832541267836128736..., but also small numbers like 10, so if you have a parent-children relationship between these two, then it can potentially overlap with your other child. Therefore, we need to first go through the children, make sure we get how much space they are having, then we use that information to construct ourself.

def __str__(self):
 h = self.getHeight()
 rowsStrs = ["" for i in range(2 * h - 1)]
 
 # return of helper is [leftLen, curLen, rightLen] where
 # leftLen = children length of left side
 # curLen = length of keyStr + length of "_" from both left side and right side
 # rightLen = children length of right side.
 # But the point of helper is to construct rowsStrs so we get the representation
 # of this BST.
 def helper(node, curRow, curCol):
 if(not node): return [0, 0, 0]
 keyStr = str(node.key)
 keyStrLen = len(keyStr)
 l = helper(node.l, curRow + 2, curCol)
 rowsStrs[curRow] += (curCol -len(rowsStrs[curRow]) + l[0] + l[1] + 1) * " " + keyStr
 if(keyStrLen < l[2] and (node.r or (node.p and node.p.l == node))): 
 rowsStrs[curRow] += (l[2] - keyStrLen) * "_"
 if(l[1]): 
 rowsStrs[curRow + 1] += (len(rowsStrs[curRow + 2]) - len(rowsStrs[curRow + 1])) * " " + "/"
 r = helper(node.r, curRow + 2, len(rowsStrs[curRow]) + 1)
 rowsStrs[curRow] += r[0] * "_"
 if(r[1]): 
 rowsStrs[curRow + 1] += (len(rowsStrs[curRow]) - len(rowsStrs[curRow + 1])) * " " + "\\"
 return [l[0] + l[1] + 1, max(l[2] - keyStrLen, 0) + keyStrLen + r[0], r[1] + r[2] + 1]
 helper(self.head, 0, 0)
 res = "\n".join(rowsStrs)
 #print("\n\n\nStart of BST:****************************************")
 #print(res)
 #print("End of BST:****************************************")
 #print("BST height: ", h, ", BST size: ", self.size)
 return res

Here's some examples of running this:

[26883404633, 10850198033, 89739221773, 65799970852, 6118714998, 31883432186, 84275473611, 25958013736, 92141734773, 91725885198, 131191476, 81453208197, 41559969292, 90704113213, 6886252839]
 26883404633___________________________________________
 / \
 10850198033__ 89739221773___________________________
 / \ / \
 6118714998_ 25958013736 65799970852_______________ 92141734773
 / \ / \ /
 131191476 6886252839 31883432186_ 84275473611 91725885198
 \ / /
 41559969292 81453208197 90704113213

Another example:

['rtqejfxpwmggfro', 'viwmdmpedzwvvxalr', 'mvvjmkdcdpcfb', 'ykqehfqbpcjfd', 'iuuujkmdcle', 'nzjbyuvlodahlpozxsc', 'wdjtqoygcgbt', 'aejduciizj', 'gzcllygjekujzcovv', 'naeivrsrfhzzfuirq', 'lwhcjbmcfmrsnwflezxx', 'gjdxphkpfmr', 'nartcxpqqongr', 'pzstcbohbrb', 'ykcvidwmouiuz']
 rtqejfxpwmggfro____________________
 / \
 mvvjmkdcdpcfb_____________________________ viwmdmpedzwvvxalr_______________
 / \ \
 iuuujkmdcle_________ nzjbyuvlodahlpozxsc_ ykqehfqbpcjfd
 / \ / \ /
 aejduciizj_____________ lwhcjbmcfmrsnwflezxx naeivrsrfhzzfuirq_ pzstcbohbrb wdjtqoygcgbt_
 \ \ \
 gzcllygjekujzcovv nartcxpqqongr ykcvidwmouiuz
 /
 gjdxphkpfmr

Answer 14

Record Each Level Separately using Breadth First Approach

You can use a breadth first traversal and record node values in a dictionary using level as key. This helps next when you want to print each level in a new line. If you maintain a count of nodes processed, you can find current node's level (since it's a binary tree) using - level = math.ceil(math.log(count + 1, 2) - 1)

Sample Code

Here's my code using the above method (along with some helpful variables like point_span & line_space which you can modify as you like). I used my custom Queue class, but you can also use a list for maintaining queue.

def pretty_print(self):
 q, current, count, level, data = Queue(), self.root, 1, 0, {}
 while current:
 level = math.ceil(math.log(count + 1, 2) - 1)
 if data.get(level) is None:
 data[level] = []
 data[level].append(current.value)
 count += 1
 if current.left:
 q.enqueue(current.left)
 if current.right:
 q.enqueue(current.right)
 current = q.dequeue()
 point_span, line_space = 8, 4
 line_width = int(point_span * math.pow(2, level))
 for l in range(level + 1):
 current, string = data[l], ''
 for c in current:
 string += str(c).center(line_width // len(current))
 print(string + '\n' * line_space)

And here's how the output looks: Pretty Print a Binary Tree

Answer 15

Below python code will do a level order traversal of the tree, and fill up a matrix that can be used for printing to the terminal. Using the matrix allows neat features like compacting the string representation of the tree, left aligning it etc; This may not be a very space efficient solution, but its nice for pretty printing. (note that if length of node values/data are different the alignments will suffer, it can be fixed with little more logic)

 2 
┌───────────────┴───────────────┐ 
1 4 
 ┌───────┴───────┐ 
 3 5 
 └───┐ 
 6 
 └─┐
 7

If Compact is set to True:

 2 
┌┴─┐ 
1 4 
 ┌┴┐ 
 3 5 
 └┐ 
 6 
 └┐
 7

Python Code

def height(node):
 if node is None:
 return 0
 left_height = height(node.left)
 right_height = height(node.right)
 return max(left_height, right_height) + 1
def display_tree(root, compact=False):
 # Calculate the height of the tree
 h = height(root)
 # Queue to store nodes at each level
 matrix = [[' '] * (2 ** h) * 2 for _ in range(h * 2)]
 col_idx = 2 ** h
 levels = [[(root, col_idx)]]
 for l in range(h):
 curr_lvl = levels[l]
 next_lvl = []
 for node, col_idx in curr_lvl:
 matrix[l * 2][col_idx] = str(node.data)
 conn_row = matrix[l * 2 + 1]
 if node.left:
 lft_idx = col_idx - 2 ** (h - l - 1)
 next_lvl.append((node.left, lft_idx))
 # connector row for children
 conn_row[col_idx] = "┘"
 conn_row[lft_idx] = "┌"
 for j in range(lft_idx + 1, col_idx):
 conn_row[j] = "─"
 if node.right:
 rt_idx = col_idx + 2 ** (h - l - 1)
 next_lvl.append((node.right, rt_idx))
 conn_row[col_idx] = "└"
 conn_row[rt_idx] = "┐"
 for j in range(col_idx + 1, rt_idx):
 conn_row[j] = "─"
 if node.left and node.right:
 conn_row[col_idx] = "┴"
 levels.append(next_lvl)
 left_align(matrix, compact)
 for row in matrix:
 print(''.join(row))
def left_align(matrix, compact=False):
 # Find the index of the first non-empty column
 empty_columns = []
 for col_idx in range(len(matrix[0])):
 for row_idx in range(len(matrix)):
 symbol = matrix[row_idx][col_idx]
 if symbol == ' ' or (symbol == '─' if compact else False):
 continue
 else:
 break
 else:
 empty_columns.append(col_idx)
 # Replace space characters with empty strings in empty columns
 for row_idx in range(len(matrix)):
 for col_idx in empty_columns:
 matrix[row_idx][col_idx] = ''
 return matrix

Answer 16

I enhanced yozn's answer to make the output more friendly.

python3 code:

from typing import Union, Optional
class TreeNode():
 def __init__(self, value: Union[int, str], left=None, right=None):
 self.value = value
 self.left = left
 self.right = right
 def draw(self, level=0, arrow: str = '↓', height: int = 3, width: int = 2) -> Optional[list]:
 _show = []
 _show += self.left.draw(level + 1, '↙', height) if self.left else []
 _show.append('{}{}{}'.format(' ' * height * level, arrow, self.value))
 _show += self.right.draw(level + 1, '↘', height) if self.right else []
 if level == 0:
 _max = len(max(_show, key=len)) if _show else 0
 for i in range(_max):
 print((' ' * width).join([x.ljust(_max, ' ')[i] for x in _show]))
 else:
 return _show
if __name__ == '__main__':
 t1 = TreeNode('A', TreeNode('B', TreeNode('D'), TreeNode('E')),
 TreeNode('C', TreeNode('F', None, TreeNode('H')), TreeNode('G')))
 t1.draw()
 print('############################################')
 t2 = TreeNode('A', TreeNode('BB', TreeNode('DDD'), TreeNode('EEE')),
 TreeNode('CC', TreeNode('FFF', None, TreeNode('H')), TreeNode('G')))
 t2.draw(height=4)

result:

 ↓ 
 A 
 
 ↙ ↘ 
 B C 
 
↙ ↘ ↙ ↘
D E F G
 
 ↘ 
 H 
############################################
 ↓ 
 A 
 
 
 ↙ ↘ 
 B C 
 B C 
 
↙ ↘ ↙ ↘
D E F G
D E F 
D E F 
 ↘ 
 H 

Answer 17

Similar question is being answered over here This may help following code will print in this format

>>> 
1
2 3
4 5 6
7
>>> 

Code for this is as below :

class Node(object):
 def __init__(self, value, left=None, right=None):
 self.value = value
 self.left = left
 self.right = right
def traverse(rootnode):
 thislevel = [rootnode]
 a = ' '
 while thislevel:
 nextlevel = list()
 a = a[:len(a)/2]
 for n in thislevel:
 print a+str(n.value),
 if n.left: nextlevel.append(n.left)
 if n.right: nextlevel.append(n.right)
 print
 thislevel = nextlevel
t = Node(1, Node(2, Node(4, Node(7)),Node(9)), Node(3, Node(5), Node(6)))
traverse(t)

Edited code gives result in this format :

>>> 
 1
 2 3
 4 9 5 6
7
>>> 

This is just a trick way to do what you want their maybe a proper method for that I suggest you to dig more into it.

• user2762315: thanks for the answer but i want to print my tree in the format i specified above(hierarchial structure)
• Prashant Shukla: I've edited my original answer , hope this solves your purpose.
• alelom: This is a copy-pasted answer from the linked question, and it does not even compile in Python, as it is written in C. Downvoted.

Answer 18

pip install vicutils

from vicutils.printBin import NodeToString
print(NodeToString(node))

node just needs to have .left and .right defined

Example

from vicutils.printBin import BinaryNode, nodeToString
lst = list(range(2 ** 4 - 1))
node = BinaryNode(lst)
print(nodeToString(node))
print("\n")
print(nodeToString(node, valueFillChar=" "))
#There are many facultative args to customize

Output

 __________0__________ 
 / \ 
 ____1____ ____2____ 
 / \ / \ 
 _3_ _4_ _5_ _6_ 
 / \ / \ / \ / \ 
_7_ _8_ _9_ _10 _11 _12 _13 _14
 _________ 0 _________ 
 / \ 
 ___ 1 ___ ___ 2 ___ 
 / \ / \ 
 3 4 5 6 
 / \ / \ / \ / \ 
 7 8 9 10 11 12 13 14