LeetCode.563 Binary Tree Tilt

最新推荐文章于 2019-03-02 23:18:29 发布

xiakexiaohu

最新推荐文章于 2019-03-02 23:18:29 发布

阅读量196

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode算法编程

本文链接：https://blog.youkuaiyun.com/xiakexiaohu/article/details/79133058

LeetCode算法编程专栏收录该内容

225 篇文章

订阅专栏

本文介绍了一种计算二叉树中所有节点的倾斜值的方法。倾斜值定义为每个节点的左子树节点值之和与右子树节点值之和的绝对差，最终返回整棵树所有节点倾斜值的总和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

分析（原创）：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int sum=0;
    
    public int findTilt(TreeNode root) {
        //给定二叉树，返回各节点的左右子树之差的绝对值的之和
        //思路：更改root原来的值，递归求解各节点的绝对差值,利用全局变量存储
        backtrace(root);
        return sum;
    }
    public void backtrace(TreeNode root){
        if(root==null) return ;
        backtrace(root.left);
        backtrace(root.right);
        //最后求解根
        root.val+=(root.left==null?0:root.left.val)+(root.right==null?0:root.right.val);
        sum+=Math.abs((root.left==null?0:root.left.val)-(root.right==null?0:root.right.val));
        
    }
    
}