Leetcode 1038. Convert binary search tree to accumulative tree

Leetcode 1038. Convert binary search tree to accumulative tree

topic

给出二叉 搜索 树的根节点，该树的节点值各不相同，请你将其转换为累加树（Greater Sum Tree），使每个节点 node 的新值等于原树中大于或等于 node.val 的值之和。

提醒一下，二叉搜索树满足下列约束条件：

节点的左子树仅包含键 小于 节点键的节点。
节点的右子树仅包含键 大于 节点键的节点。
左右子树也必须是二叉搜索树。
注意：该题目与 538: https://leetcode-cn.com/problems/convert-bst-to-greater-tree/  相同

Example 1:

输入：[4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
输出：[30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Example 2:

输入：root = [0,null,1]
输出：[1,null,1]

Example 3:

输入：root = [1,0,2]
输出：[3,3,2]

Example 4:

输入：root = [3,2,4,1]
输出：[7,9,4,10]

Ideas

• Basically, the tree problem can be handled by recursion, this problem is also the same idea
• Since the in-order traversal of the binary search tree meets the characteristics of increasing, this problem can be considered using in-order traversal
• If it is left->root->right, you will find that it is not easy to write, but if you change it to the order of right->root->left, when you traverse to root, you will have all the nodes greater than or equal to root->val It's all over

Code

/**
 * 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:
    void dfs(TreeNode* root, int& sum) {
        if (!root) return;

        dfs(root->right, sum);
        sum += root->val;
        root->val = sum;
        dfs(root->left, sum);
    }

    TreeNode* bstToGst(TreeNode* root) {
        int sum = 0;
        dfs(root, sum);
        return root;
    }
};