Binary Tree Tilt - LintCode

最新推荐文章于 2019-08-21 09:12:47 发布

zhaokane

最新推荐文章于 2019-08-21 09:12:47 发布

阅读量218

点赞数

CC 4.0 BY-SA版权

分类专栏： LintCode C++

本文链接：https://blog.youkuaiyun.com/zhaohengchuan/article/details/80867545

LintCode 同时被 2 个专栏收录

428 篇文章

订阅专栏

C++

204 篇文章

订阅专栏

本文介绍了一种计算二叉树节点倾斜值的方法，通过递归遍历将每个节点的值更新为其左右子树值之和，并计算左右子树值之差的绝对值作为该节点的倾斜值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

描述
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.

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

样例

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

思路
将二叉树节点的值置为左右子树的值得和,节点的tilt就是左右子树值差的绝对值。

#ifndef C1172_H
#define C1172_H
#include<iostream>
#include<cmath>
using namespace std;
class TreeNode{
public:
    int val;
    TreeNode *left, *right;
    TreeNode(int val){
        this->val = val;
        this->left = this->right = NULL;
    }
};
class Solution {
public:
    /**
    * @param root: the root
    * @return: the tilt of the whole tree
    */
    int findTilt(TreeNode * root) {
        // Write your code here
        if (!root)
            return 0;
        nodeSum(root);
        return tilt;
    }
    //将结点的值置为左右子树值的和
    int nodeSum(TreeNode *root)
    {
        if (!root)
            return 0;
        if (!root->left&&!root->right)
            return root->val;
        root->val += nodeSum(root->left);
        root->val += nodeSum(root->right);
        int l = 0, r = 0;
        if (root->left)
            l = root->left->val;
        if (root->right)
            r = root->right->val;
        tilt += abs(l - r);//计算结点的tilt值
        return root->val;
    }
    int tilt = 0;//记录所有结点tilt值得和
};
#endif