LeetCode-538. Convert BST to Greater TreeLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/convert-bst-to-greater-tree/

Given the root 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 the sum of all keys greater than the original key in BST.

As a reminder, a binary search tree is a tree that satisfies these constraints:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both 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]

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -104 <= Node.val <= 104
• All the values in the tree are unique.
• root is guaranteed to be a valid binary search tree.

Note: This question is the same as 1038: LeetCode-1038. Binary Search Tree to Greater Sum TreeLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

【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) {
         modify(root,0);
        return root;
    }
    int modify(TreeNode* root, int val){
        if (!root) return val;
        int r = modify(root->right, val);
        root->val += r;
        return modify(root->left,root->val);
    }
};

【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) {
        modify(root,0);
        return root;
    }
    int modify(TreeNode root, int val){
        if (root == null) {return val;}
        int r = modify(root.right, val);
        root.val += r;
        return modify(root.left, root.val);
    }
}