Leetcode 538. Convert BST to Greater Tree (反向inorder遍历)

  1. Convert BST to Greater Tree
    Medium
    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: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

解法1：用反向的inorder排序。

/**
 * 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) {
        helper(root);
        return root;        
    }
private:
    int sum;
    void helper(TreeNode* root) {
        if (!root) return;
        helper(root->right);
        sum += root->val;
        root->val = sum;
        helper(root->left);
    }
};