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# Sword finger Offer II 054. Sum of all values greater than or equal to nodes ｜ 538. Convert binary search tree to accumulation tree ｜ 1038. Convert binary search tree to accumulation tree:

Given a binary search tree, replace the value of each node with the sum of all node values greater than or equal to the node value in the tree.

Remind me that the binary search tree meets the following constraints:

- The left subtree of a node contains only nodes whose key is less than the node key.
- The right subtree of a node contains only nodes whose key is greater than the node key.
- The left and right subtrees must also be binary search trees.

# Example 1

Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8] Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

# Example 2

Input: root = [0,null,1] Output: [1,null,1]

# Example 3

Input: root = [1,0,2] Output: [3,3,2]

# Example 4

Input: root = [3,2,4,1] Output: [7,9,4,10]

# Tips

- The number of nodes in the tree is between 0 and 104.
- The value of each node is between - 104 and 104.
- All values in the tree are different from each other.
- The given tree is a binary search tree.

# analysis

- This algorithm problem needs to be translated and easy to understand.
- Binary tree, you should be familiar with, is generally positive order traversal, medium order traversal, post order traversal, and then some simple logic in the process of traversal.
- Binary search tree is a special binary tree. The problem has been explained.
- Because the value of each node should be replaced by the sum of all node values greater than or equal to the node value in the tree, according to the characteristics of binary search tree, in fact, it is to replace the value of each node with its own value and accumulate the sum of the values of the right subtree.
- First of all, you need to traverse each node, but the order is the key. You will find that the normal positive order traversal, medium order traversal and post order traversal are not smooth.
- According to the meaning of the topic, it is obvious that the best order is to traverse the right subtree, then the root node, and then the left subtree. This is the most smooth. Only one variable needs to continuously accumulate the node value, accumulate the value of which node to traverse, and then assign it to this node. This is just the opposite of the middle order traversal, that is, the reverse middle order traversal.
- Traversing this doll structure is generally recursion or loop (using stack data structure). Recursion is relatively simple, but limited by the call stack. Loop actually simulates the call stack with a data structure such as stack, which is basically the same, but the number of calls can be more, because the space of the call stack is generally limited and the heap space is much larger, However, to achieve the same function, it is logically more complex.
- Whether you use recursion or stack data structure to cycle, you need additional memory space corresponding to the tree structure. There is a very clever traversal method called Morris traversal, which can reduce the space complexity of non recursive traversal to O (1). The specific implementation of the second leader has been annotated in detail. You can also go to the Encyclopedia for in-depth understanding.

# Problem solution

## java

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public TreeNode convertBST(TreeNode root) { int sum = 0; TreeNode cur = root; while (cur != null) { if (cur.right == null) { // No child is worth more than himself // It's your turn to deal with yourself sum += cur.val; cur.val = sum; // I have traversed it myself, so I should traverse a smaller one than myself cur = cur.left; } else { // On the far left of the right child, the youngest child older than himself // That is, the last traversal node in front of you, and the next one should be yourself TreeNode leftmostOfRight = cur.right; while (leftmostOfRight.left != null && leftmostOfRight.left != cur) { leftmostOfRight = leftmostOfRight.left; } if (leftmostOfRight.left == null) { // No association has been done. It means that the current node is associated for the first time to ensure that it is your turn after traversing the smallest node larger than yourself leftmostOfRight.left = cur; cur = cur.right; } else { // The second traversal here shows that the larger ones are traversed // Dissolve the relationship and restore the tree structure leftmostOfRight.left = null; // It's your turn to deal with yourself sum += cur.val; cur.val = sum; // I have traversed it myself, so I should traverse a smaller one than myself cur = cur.left; } } } return root; } }

## c

/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ struct TreeNode* convertBST(struct TreeNode* root){ int sum = 0; struct TreeNode *cur = root; while (cur != NULL) { if (cur->right == NULL) { // No child is worth more than himself // It's your turn to deal with yourself sum += cur->val; cur->val = sum; // I have traversed it myself, so I should traverse a smaller one than myself cur = cur->left; } else { // On the far left of the right child, the youngest child older than himself // That is, the last traversal node in front of you, and the next one should be yourself struct TreeNode *leftmostOfRight = cur->right; while (leftmostOfRight->left != NULL && leftmostOfRight->left != cur) { leftmostOfRight = leftmostOfRight->left; } if (leftmostOfRight->left == NULL) { // No association has been done. It means that the current node is associated for the first time to ensure that it is your turn after traversing the smallest node larger than yourself leftmostOfRight->left = cur; cur = cur->right; } else { // The second traversal here shows that the larger ones are traversed // Dissolve the relationship and restore the tree structure leftmostOfRight->left = NULL; // It's your turn to deal with yourself sum += cur->val; cur->val = sum; // I have traversed it myself, so I should traverse a smaller one than myself cur = cur->left; } } } return root; }

## c++

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: TreeNode* convertBST(TreeNode* root) { int sum = 0; TreeNode *cur = root; while (cur != nullptr) { if (cur->right == nullptr) { // No child is worth more than himself // It's your turn to deal with yourself sum += cur->val; cur->val = sum; // I have traversed it myself, so I should traverse a smaller one than myself cur = cur->left; } else { // On the far left of the right child, the youngest child older than himself // That is, the last traversal node in front of you, and the next one should be yourself TreeNode *leftmostOfRight = cur->right; while (leftmostOfRight->left != nullptr && leftmostOfRight->left != cur) { leftmostOfRight = leftmostOfRight->left; } if (leftmostOfRight->left == nullptr) { // No association has been done. It means that the current node is associated for the first time to ensure that it is your turn after traversing the smallest node larger than yourself leftmostOfRight->left = cur; cur = cur->right; } else { // The second traversal here shows that the larger ones are traversed // Dissolve the relationship and restore the tree structure leftmostOfRight->left = nullptr; // It's your turn to deal with yourself sum += cur->val; cur->val = sum; // I have traversed it myself, so I should traverse a smaller one than myself cur = cur->left; } } } return root; } };

## python

# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def convertBST(self, root: TreeNode) -> TreeNode: total = 0 cur = root while cur: if not cur.right: # No child is worth more than himself # It's your turn to deal with yourself total += cur.val cur.val = total # I have traversed it myself, so I should traverse a smaller one than myself cur = cur.left else: # On the far left of the right child, the youngest child older than himself # That is, the last traversal node in front of you, and the next one should be yourself leftmostOfRight = cur.right while leftmostOfRight.left and leftmostOfRight.left != cur: leftmostOfRight = leftmostOfRight.left if not leftmostOfRight.left: # No association has been done. It means that the current node is associated for the first time to ensure that it is your turn after traversing the smallest node larger than yourself leftmostOfRight.left = cur cur = cur.right else: # The second traversal here shows that the larger ones are traversed # Dissolve the relationship and restore the tree structure leftmostOfRight.left = None # It's your turn to deal with yourself total += cur.val cur.val = total # I have traversed it myself, so I should traverse a smaller one than myself cur = cur.left return root

## go

/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func convertBST(root *TreeNode) *TreeNode { sum := 0 cur := root for cur != nil { if cur.Right == nil { // No child is worth more than himself // It's your turn to deal with yourself sum += cur.Val cur.Val = sum // I have traversed it myself, so I should traverse a smaller one than myself cur = cur.Left } else { // On the far left of the right child, the youngest child older than himself // That is, the last traversal node in front of you, and the next one should be yourself leftmostOfRight := cur.Right for leftmostOfRight.Left != nil && leftmostOfRight.Left != cur { leftmostOfRight = leftmostOfRight.Left } if leftmostOfRight.Left == nil { // No association has been done. It means that the current node is associated for the first time to ensure that it is your turn after traversing the smallest node larger than yourself leftmostOfRight.Left = cur cur = cur.Right } else { // The second traversal here shows that the larger ones are traversed // Dissolve the relationship and restore the tree structure leftmostOfRight.Left = nil // It's your turn to deal with yourself sum += cur.Val cur.Val = sum // I have traversed it myself, so I should traverse a smaller one than myself cur = cur.Left } } } return root }

## rust

// Definition for a binary tree node. // #[derive(Debug, PartialEq, Eq)] // pub struct TreeNode { // pub val: i32, // pub left: Option<Rc<RefCell<TreeNode>>>, // pub right: Option<Rc<RefCell<TreeNode>>>, // } // // impl TreeNode { // #[inline] // pub fn new(val: i32) -> Self { // TreeNode { // val, // left: None, // right: None // } // } // } use std::rc::Rc; use std::cell::RefCell; impl Solution { pub fn convert_bst(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> { let mut sum = 0; let mut cur = root.clone(); while cur.is_some() { if cur.as_ref().unwrap().borrow().right.is_none() { // No child is worth more than himself // It's your turn to deal with yourself sum += cur.as_ref().unwrap().borrow().val; cur.as_ref().unwrap().borrow_mut().val = sum; // I have traversed it myself, so I should traverse a smaller one than myself cur = cur.unwrap().borrow().left.clone(); } else { // On the far left of the right child, the youngest child older than himself // That is, the last traversal node in front of you, and the next one should be yourself let mut leftmostOfRight = cur.as_ref().unwrap().borrow().right.clone(); while leftmostOfRight.as_ref().unwrap().borrow().left.is_some() && leftmostOfRight.as_ref().unwrap().borrow().left != cur { leftmostOfRight = leftmostOfRight.unwrap().borrow().left.clone(); } if leftmostOfRight.as_ref().unwrap().borrow().left.is_none() { // No association has been done. It means that the current node is associated for the first time to ensure that it is your turn after traversing the smallest node larger than yourself leftmostOfRight.unwrap().borrow_mut().left = cur.clone(); cur = cur.unwrap().borrow().right.clone(); } else { // The second traversal here shows that the larger ones are traversed // Dissolve the relationship and restore the tree structure leftmostOfRight.unwrap().borrow_mut().left = Option::None; // It's your turn to deal with yourself sum += cur.as_ref().unwrap().borrow().val; cur.as_ref().unwrap().borrow_mut().val = sum; // I have traversed it myself, so I should traverse a smaller one than myself cur = cur.unwrap().borrow().left.clone(); } } } root } }