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Commit ef7e407

Browse files
author
liningrui
committed
add new solutions
1 parent 579ed13 commit ef7e407

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+2044
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‎script/[100]相同的树.py

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# 给你两棵二叉树的根节点 p 和 q ,编写一个函数来检验这两棵树是否相同。
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#
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# 如果两个树在结构上相同,并且节点具有相同的值,则认为它们是相同的。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入:p = [1,2,3], q = [1,2,3]
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# 输出:true
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#
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#
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# 示例 2:
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#
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#
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# 输入:p = [1,2], q = [1,null,2]
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# 输出:false
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#
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#
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# 示例 3:
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#
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#
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# 输入:p = [1,2,1], q = [1,1,2]
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# 输出:false
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#
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#
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#
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#
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# 提示:
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#
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#
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# 两棵树上的节点数目都在范围 [0, 100] 内
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# -104 <= Node.val <= 104
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#
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# Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 862 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
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def helper(p,q):
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if not p and not q:
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return True
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elif not p:
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return False
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elif not q:
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return False
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elif p.val!=q.val:
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return False
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return helper(p.right, q.right) and helper(p.left, q.left)
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return helper(p,q)
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[101]对称二叉树.py

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# 给你一个二叉树的根节点 root , 检查它是否轴对称。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入:root = [1,2,2,3,4,4,3]
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# 输出:true
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#
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#
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# 示例 2:
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#
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#
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# 输入:root = [1,2,2,null,3,null,3]
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# 输出:false
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#
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#
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#
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#
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# 提示:
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#
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#
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# 树中节点数目在范围 [1, 1000] 内
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# -100 <= Node.val <= 100
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#
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#
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#
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#
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# 进阶:你可以运用递归和迭代两种方法解决这个问题吗?
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# Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 2013 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def isSymmetric(self, root: Optional[TreeNode]) -> bool:
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def helper(p,q):
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if not p and not q:
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return True
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elif not p:
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return False
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elif not q:
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return False
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elif p.val!=q.val:
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return False
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return helper(p.left, q.right) and helper(p.right, q.left)
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return helper(root.left, root.right)
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[102]二叉树的层序遍历.py

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# 给你二叉树的根节点 root ,返回其节点值的 层序遍历 。 (即逐层地,从左到右访问所有节点)。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入:root = [3,9,20,null,null,15,7]
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# 输出:[[3],[9,20],[15,7]]
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#
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#
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# 示例 2:
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#
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#
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# 输入:root = [1]
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# 输出:[[1]]
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#
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#
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# 示例 3:
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#
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#
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# 输入:root = []
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# 输出:[]
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#
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#
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#
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#
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# 提示:
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#
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#
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# 树中节点数目在范围 [0, 2000] 内
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# -1000 <= Node.val <= 1000
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#
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# Related Topics 树 广度优先搜索 二叉树 👍 1391 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def levelOrder(self, root: TreeNode) -> List[List[int]]:
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if not root:
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return []
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result = []
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stack = [root]
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while stack:
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l = len(stack)
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tmp = []
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for i in range(l):
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node= stack.pop(0) #注意是pop left
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tmp.append(node.val)
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if node.left:
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stack.append(node.left)
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if node.right:
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stack.append(node.right)
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result.append(tmp)
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return result
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# leetcode submit region end(Prohibit modification and deletion)
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# 给你二叉树的根节点 root ,返回其节点值的 锯齿形层序遍历 。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入:root = [3,9,20,null,null,15,7]
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# 输出:[[3],[20,9],[15,7]]
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#
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#
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# 示例 2:
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#
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#
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# 输入:root = [1]
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# 输出:[[1]]
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#
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#
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# 示例 3:
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#
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#
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# 输入:root = []
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# 输出:[]
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#
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#
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#
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#
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# 提示:
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#
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#
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# 树中节点数目在范围 [0, 2000] 内
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# -100 <= Node.val <= 100
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#
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# Related Topics 树 广度优先搜索 二叉树 👍 668 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
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if not root:
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return []
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result = []
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stack = [root]
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flag = 1
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while stack:
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l = len(stack)
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tmp = []
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for i in range(l):
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node= stack.pop(0) #注意是pop left
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tmp.append(node.val)
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if node.left:
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stack.append(node.left)
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if node.right:
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stack.append(node.right)
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if flag>0:
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result.append(tmp)
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else:
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result.append(tmp[::-1])
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flag*=-1
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return result
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[104]二叉树的最大深度.py

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# 给定一个二叉树,找出其最大深度。
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#
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# 二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。
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#
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# 说明: 叶子节点是指没有子节点的节点。
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#
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# 示例:
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# 给定二叉树 [3,9,20,null,null,15,7],
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#
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# 3
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# / \
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# 9 20
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# / \
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# 15 7
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#
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# 返回它的最大深度 3 。
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# Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 1300 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def maxDepth(self, root: Optional[TreeNode]) -> int:
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def dfs(root):
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if not root:
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return 0
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left= dfs(root.left)
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right = dfs(root.right)
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return max(left,right)+1
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return dfs(root)
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# leetcode submit region end(Prohibit modification and deletion)
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# 给定两个整数数组 preorder 和 inorder ,其中 preorder 是二叉树的先序遍历, inorder 是同一棵树的中序遍历,请构造二叉树并
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# 返回其根节点。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
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# 输出: [3,9,20,null,null,15,7]
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#
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#
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# 示例 2:
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#
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#
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# 输入: preorder = [-1], inorder = [-1]
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# 输出: [-1]
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#
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#
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#
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#
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# 提示:
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#
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#
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# 1 <= preorder.length <= 3000
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# inorder.length == preorder.length
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# -3000 <= preorder[i], inorder[i] <= 3000
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# preorder 和 inorder 均 无重复 元素
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# inorder 均出现在 preorder
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# preorder 保证 为二叉树的前序遍历序列
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# inorder 保证 为二叉树的中序遍历序列
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#
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# Related Topics 树 数组 哈希表 分治 二叉树 👍 1663 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
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def dfs(left, right):
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if left>right:
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return None
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val = preorder.pop(0)
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root = TreeNode(val)
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pos = inorder.index(val)
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root.left = dfs(left, pos-1)
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root.right = dfs(pos+1, right)
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return root
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return dfs(0, len(preorder)-1)
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# leetcode submit region end(Prohibit modification and deletion)

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