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Commit f85d9a0

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feat: add solutions to lc problems: No.0538,1038
1 parent b74ee20 commit f85d9a0

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-89
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11 files changed

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‎solution/0500-0599/0538.Convert BST to Greater Tree/README.md‎

Lines changed: 44 additions & 2 deletions
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@@ -57,7 +57,6 @@
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<li>给定的树为二叉搜索树。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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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
77+
class Solution:
78+
add = 0
79+
def convertBST(self, root: TreeNode) -> TreeNode:
80+
if root:
81+
self.convertBST(root.right)
82+
root.val += self.add
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self.add = root.val
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self.convertBST(root.left)
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return root
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```
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### **Java**
@@ -91,6 +104,35 @@ class Solution {
91104
}
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```
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### **C++**
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
120+
*/
121+
class Solution {
122+
public:
123+
int add = 0;
124+
TreeNode* convertBST(TreeNode* root) {
125+
if (root) {
126+
convertBST(root->right);
127+
root->val += add;
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add = root->val;
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convertBST(root->left);
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}
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return root;
132+
}
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};
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```
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94136
### **...**
95137
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```

‎solution/0500-0599/0538.Convert BST to Greater Tree/README_EN.md‎

Lines changed: 71 additions & 25 deletions
Original file line numberDiff line numberDiff line change
@@ -6,24 +6,16 @@
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<p>Given the <code>root</code> of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.</p>
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<p>As a reminder, a <em>binary search tree</em> is a tree that satisfies these constraints:</p>
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<ul>
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<li>The left subtree of a node contains only nodes with keys&nbsp;<strong>less than</strong>&nbsp;the node&#39;s key.</li>
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<li>The right subtree of a node contains only nodes with keys&nbsp;<strong>greater than</strong>&nbsp;the node&#39;s key.</li>
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<li>Both the left and right subtrees must also be binary search trees.</li>
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</ul>
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<p><strong>Note:</strong> This question is the same as&nbsp;1038:&nbsp;<a href="https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/">https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/</a></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> root = [0,null,1]
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> root = [1,0,2]
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</pre>
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<p><strong>Example 4:</strong></p>
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<pre>
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<strong>Input:</strong> root = [3,2,4,1]
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
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<li><code>-10<sup>4</sup> &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
@@ -102,7 +78,21 @@
10278
### **Python3**
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```python
105-
81+
# Definition for a binary tree node.
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# class TreeNode:
83+
# def __init__(self, val=0, left=None, right=None):
84+
# self.val = val
85+
# self.left = left
86+
# self.right = right
87+
class Solution:
88+
add = 0
89+
def convertBST(self, root: TreeNode) -> TreeNode:
90+
if root:
91+
self.convertBST(root.right)
92+
root.val += self.add
93+
self.add = root.val
94+
self.convertBST(root.left)
95+
return root
10696
```
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### **Java**
@@ -122,6 +112,62 @@ class Solution {
122112
}
123113
```
124114

115+
### **C++**
116+
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```cpp
118+
/**
119+
* Definition for a binary tree node.
120+
* struct TreeNode {
121+
* int val;
122+
* TreeNode *left;
123+
* TreeNode *right;
124+
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
125+
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
126+
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
127+
* };
128+
*/
129+
class Solution {
130+
public:
131+
int add = 0;
132+
TreeNode* convertBST(TreeNode* root) {
133+
if (root) {
134+
convertBST(root->right);
135+
root->val += add;
136+
add = root->val;
137+
convertBST(root->left);
138+
}
139+
return root;
140+
}
141+
};
142+
```
143+
144+
### **Go**
145+
146+
```go
147+
/**
148+
* Definition for a binary tree node.
149+
* type TreeNode struct {
150+
* Val int
151+
* Left *TreeNode
152+
* Right *TreeNode
153+
* }
154+
*/
155+
func convertBST(root *TreeNode) *TreeNode {
156+
add := 0
157+
var dfs func(*TreeNode)
158+
dfs = func(node *TreeNode) {
159+
if node != nil {
160+
dfs(node.Right)
161+
node.Val += add
162+
add = node.Val
163+
dfs(node.Left)
164+
}
165+
}
166+
dfs(root)
167+
return root
168+
}
169+
```
170+
125171
### **...**
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```
Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
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/**
2+
* Definition for a binary tree node.
3+
* struct TreeNode {
4+
* int val;
5+
* TreeNode *left;
6+
* TreeNode *right;
7+
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
8+
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
9+
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10+
* };
11+
*/
12+
class Solution {
13+
public:
14+
int add = 0;
15+
TreeNode* convertBST(TreeNode* root) {
16+
if (root) {
17+
convertBST(root->right);
18+
root->val += add;
19+
add = root->val;
20+
convertBST(root->left);
21+
}
22+
return root;
23+
}
24+
};
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,22 @@
1+
/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
6+
* Right *TreeNode
7+
* }
8+
*/
9+
func convertBST(root *TreeNode) *TreeNode {
10+
add := 0
11+
var dfs func(*TreeNode)
12+
dfs = func(node *TreeNode) {
13+
if node != nil {
14+
dfs(node.Right)
15+
node.Val += add
16+
add = node.Val
17+
dfs(node.Left)
18+
}
19+
}
20+
dfs(root)
21+
return root
22+
}
Lines changed: 16 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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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):
4+
# self.val = val
5+
# self.left = left
6+
# self.right = right
7+
class Solution:
8+
add = 0
9+
10+
def convertBST(self, root: TreeNode) -> TreeNode:
11+
if root:
12+
self.convertBST(root.right)
13+
root.val += self.add
14+
self.add = root.val
15+
self.convertBST(root.left)
16+
return root

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