LeetCode Binary Tree Tilt-优快云博客

本文介绍了LeetCode上的二叉树倾斜问题，采用后序遍历的方法，自下而上计算每个节点左右子树的节点值之和，并计算其差的绝对值。最终求得整棵树的倾斜值。

原题链接在这里：https://leetcode.com/problems/binary-tree-tilt/description/

题目：

Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes' tilt.

Example:

Input: 
         1
       /   \
      2     3
Output: 1
Explanation: 
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1

Note:

The sum of node values in any subtree won't exceed the range of 32-bit integer.
All the tilt values won't exceed the range of 32-bit integer.

题解：

用postorder traverse, 自下而上先算left的和，再算right的和, 取差的绝对值加进res中. postorder 返回包括该点在内的和.

Time Complexity: O(n). n 是node个数.

Space: O(logn), stack space.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     
12     int res = 0;
13     
14     public int findTilt(TreeNode root) {
15         postorder(root);
16         return res;
17     }
18     
19     private int postorder(TreeNode root){
20         if(root == null){
21             return 0;
22         }
23         
24         int left = postorder(root.left);
25         int right = postorder(root.right);
26         res += Math.abs(left - right);
27         return left+right+root.val;
28     }
29 }