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Lc1038

05 August 2008

1038 Binary Search Tree to Greater Sum Tree

题目

Given the root of a binary search tree with distinct values, modify it so that every 

1

node
has a new value equal to the sum of the values of the original tree that are greater than or equal to 
1

node.val
.

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:

1
2

Input: [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]

Constraints:

  1. The number of nodes in the tree is between 
    1

    1
    and 
    1

    100
    .
  2. Each node will have value between 
    1

    0
    and 
    1

    100
    .
  3. The given tree is a binary search tree.

Note: This question is the same as 538: https://leetcode.com/problems/convert-bst-to-greater-tree/

分析

与538完全一样。

代码

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/**
 * 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 bstToGst(TreeNode root) {
        dfs(root, 0);
        return root;
    }
    private int dfs(TreeNode node, int sum) {
        if (node == null) {
            return sum;
        }
        int right = dfs(node.right, sum); // 先计算当前节点左子树的sum
        int left = dfs(node.left, node.val + right); // 递归右子树的值
        
        node.val += right; //当前节点需要加上右子树的值
        
        return left; // 最后遍历的地方
    }
}
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