chore: add daily leetcode post 538.把二叉搜索树转换为累加树_translated

Author: longsizhuo

app/docs/CommunityShare/Leetcode/538.把二叉搜索树转换为累加树_translated.md

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+---
+title: 538.Convert the binary search tree to cumulative tree
+date: '2024.01.01 0:00'
+tags:
+  - - Python
+  - - answer
+  - - Binary tree
+abbrlink: 32401b69
+---
+
+# topic：
+
+"""
+
+<p>Give two forks<strong> search </strong>Root node，The node value of the tree is different，Please convert it to a cumulative tree（Greater Sum Tree），Make each node <code>node</code>&nbsp;The new value of the original tree is greater than or equal to the original tree&nbsp;<code>node.val</code>&nbsp;The sum of the value。</p>
+
+<p>remind，二叉searchTree满足下列约束条件：</p>
+
+<ul> 
+ <li>The left tree of the node contains only the key<strong> Less than </strong>Node node node。</li> 
+ <li>The right tree of the node contains only the key<strong> more than the</strong> Node node node。</li> 
+ <li>左右子Tree也必须是二叉searchTree。</li> 
+</ul>
+
+<p><strong>Notice：</strong>This question 1038:&nbsp;<a href="https://leetcode-cn.com/problems/binary-search-tree-to-greater-sum-tree/">https://leetcode-cn.com/problems/binary-search-tree-to-greater-sum-tree/</a> same</p>
+
+<p>&nbsp;</p>
+
+<p><strong>Exemplary example 1：</strong></p>
+
+<p><strong><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/05/03/tree.png" style="height: 364px; width: 534px;" /></strong></p>
+
+<pre><strong>enter：</strong>[4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
+<strong>Output：</strong>[30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
+</pre>
+
+<p><strong>Exemplary example 2：</strong></p>
+
+<pre><strong>enter：</strong>root = [0,null,1]
+<strong>Output：</strong>[1,null,1]
+</pre>
+
+<p><strong>Exemplary example 3：</strong></p>
+
+<pre><strong>enter：</strong>root = [1,0,2]
+<strong>Output：</strong>[3,3,2]
+</pre>
+
+<p><strong>Exemplary example 4：</strong></p>
+
+<pre><strong>enter：</strong>root = [3,2,4,1]
+<strong>Output：</strong>[7,9,4,10]
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>hint：</strong></p>
+
+<ul> 
+ <li>The number of nodes in the tree is in <code>0</code>&nbsp;and <code>10<sup>4</sup></code><sup>&nbsp;</sup>between。</li> 
+ <li>The value of each node <code>-10<sup>4</sup></code>&nbsp;and&nbsp;<code>10<sup>4</sup></code>&nbsp;between。</li> 
+ <li>All the values ​​in the tree <strong>互不same</strong> 。</li> 
+ <li>给定的Tree为二叉searchTree。</li> 
+</ul>
+
+<div><div>Related Topics</div><div><li>Tree</li><li>深度优先search</li><li>二叉searchTree</li><li>Binary tree</li></div></div><br><div><li>👍 904</li><li>👎 0</li></div>
+"""
+
+# Thought：
+
+method one：Travel in the preface
+Thinking and algorithm
+
+本题中要求我们将每个节点的值修改为原来的节点值加上所有more than the它的节点值之and。这样我们只需要Travel in the preface该二叉searchTree，记录过程中的节点值之and，Not continuously update the node value of the nodes you currently traversed，即可得到topic要求的累加Tree。
+
+Method Two：Morris Traversal
+Thinking and algorithm
+
+There is a clever way to be online，只占用常数空间来实现中序Traversal。This method is caused by J. H. Morris exist 1979 Annual thesis「Traversing Binary Trees Simply and Cheaply」Proposed for the first time，Because it is called Morris Traversal。
+
+我们以一个简单的Binary tree为例进行说明。假设我们要进行中序Traversal的 Morris Traversal。
+
+```markdown
+          1
+         / \
+        2   3
+       / \
+      4   5
+
+```
+1. Initialization The current node is the root node（current = 1）。
+
+2. When the current node is not empty，Execute the following steps：
+
+   1. The left node of the current node is not empty。We find the front -drive node of the current node，也就是左子Tree中的最右节点。exist这个例子中，The front -drive node is the node5。
+   2. The right node of the front -drive node is empty，Point your right node to the current node（5 -> 1）。
+   3. Move the current node to its left child node（current = 2）。
+    ```markdown
+              1
+               \
+                2
+               / \
+              4   5
+    ```
+    
+    1. The left node of the current node is not empty。We find the front -drive node of the current node，也就是左子Tree中的最右节点。exist这个例子中，The front -drive node is the node4。
+    2. The right node of the front -drive node is empty，Point your right node to the current node（4 -> 2）。
+    3. Move the current node to its left child node（current = 4）。
+    ```markdown
+              1
+               \
+                2
+                 \
+                  4
+                   \
+                    5
+    
+    ```
+
+    1. The left node of the current node is empty。Output当前节点的值（4）。
+    
+    2. Move the current node to its right node（current = 5）。
+    
+    3. The left node of the current node is empty。Output当前节点的值（5）。
+    
+    4. Move the current node to its right node（current = 2）。
+    ```markdown
+              1
+               \
+                2
+                 \
+                  4
+    ```
+    1. The left node of the current node is empty。Output当前节点的值（2）。
+    2. Move the current node to its right node（current = 1）。
+    ```markdown
+              1
+               \
+                2
+    ```
+    1. The left node of the current node is empty。Output当前节点的值（1）。
+    2. Move the current node to its right node（current = 3）。
+    ```markdown
+              1
+               \
+                2
+    ```
+    1. The left node of the current node is empty。Output当前节点的值（3）。
+    2. Move the current node to its right node（current = null）。
+3. The current node is empty，Traversal结束。
+4. 
+According to the above steps，通过修改节点between的指针关系，我们完成了对Binary tree的中序Traversal。
+
+
+# Code：
+```python 中序Traversal
+class Solution:
+    def convertBST(self, root: TreeNode) -> TreeNode:
+        def dfs(root: TreeNode):
+            nonlocal total
+            if root:
+                dfs(root.right)
+                total += root.val
+                root.val = total
+                dfs(root.left)
+        
+        total = 0
+        dfs(root)
+        return root
+```
+
+```python MorrisTraversal
+class Solution:
+    def convertBST(self, root: TreeNode) -> TreeNode:
+        # Get the successor node of the given node（中序Traversal中的下一个节点）
+        def getSuccessor(node: TreeNode) -> TreeNode:
+            succ = node.right
+            while succ.left and succ.left != node:
+                succ = succ.left
+            return succ
+        
+        total = 0  # 记录累加的节点值之and
+        node = root  # The current node initialization is the root node
+
+        while node:
+            if not node.right:  # If the right node of the current node is empty
+                total += node.val  # Add the value of the current node to the cumulative value
+                node.val = total  # Update the value of the current node to accumulate value
+                node = node.left  # Move the current node to its left child node
+            else:  # If the right node of the current node is not empty
+                succ = getSuccessor(node)  # Get the successor node of the current node
+                if not succ.left:  # If the left node of the successor node is empty
+                    succ.left = node  # Point the left -node of the successor node to the current node，Establish a clue
+                    node = node.right  # Move the current node to its right node
+                else:  # If the left node of the successor node is not empty
+                    succ.left = None  # Set the left node of the successor node to empty，Remove clues
+                    total += node.val  # Add the value of the current node to the cumulative value
+                    node.val = total  # Update the value of the current node to accumulate value
+                    node = node.left  # Move the current node to its left child node
+
+        return root  # Return to the root node
+
+class Solution:
+    def convertBST(self, root: TreeNode) -> TreeNode:
+        def convert(node: TreeNode, total: int) -> int:
+            if not node:
+                return total
+
+            # 递归Traversal右子Tree
+            total = convert(node.right, total)
+
+            # Update the node value to accumulate and add value
+            total += node.val
+            node.val = total
+
+            # 递归Traversal左子Tree
+            total = convert(node.left, total)
+
+            return total
+
+        convert(root, 0)
+        return root
+```