leetcode Binary Tree Maximum Path Sum

最新推荐文章于 2024-03-25 21:39:45 发布

zyfo2

最新推荐文章于 2024-03-25 21:39:45 发布

阅读量993

点赞数

CC 4.0 BY-SA版权

分类专栏： interview questions

本文链接：https://blog.youkuaiyun.com/zyfo2/article/details/8636205

interview questions 专栏收录该内容

22 篇文章

订阅专栏

本文探讨了如何在二叉树中寻找最大路径和的问题，包括路径可以任意起点和终点的特点，并提供了一种有效的解决方案。文章通过递归方法解决了该问题，能够处理包含负数的情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

It may contain negative values. So you'll have to deal with it like the biggest subarray sum problem. the idea is exactly the same

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(!root) return 0;
        int maxs=INT_MIN;
        get(root, &maxs);
        return maxs;
    }
    
    int get(TreeNode *root, int * maxs){
        if(!root) return 0;

        int l=get(root->left,maxs);
        int r=get((*root).right,maxs);
        if(l+r+root->val>(*maxs)) *maxs=l+r+root->val;
        int t=max(l,r);
        
        return t+root->val>0?t+root->val:0;
    }
};