LeetCode-Binary Tree Maximum Path Sum

最新推荐文章于 2024-03-25 21:39:45 发布

原创最新推荐文章于 2024-03-25 21:39:45 发布 · 339 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #Binary Tree Maximum

Cracking LeetCode 同时被 3 个专栏收录

152 篇文章

订阅专栏

Tree

26 篇文章

订阅专栏

Recursion

24 篇文章

订阅专栏

本文探讨了如何在二叉树中找到最大的路径和。通过递归方法，我们能够有效地解决这个问题，理解其背后的算法逻辑。

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.

Solution:

Code:

<span style="font-size:14px;">/**
 * 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 helper(TreeNode *root, int &result) {
       if (!root) return 0;
       int left = helper(root->left, result);
       int right = helper(root->right, result);
       result = max(result, root->val+max(0, left)+max(0, right));
       left = max(left, right);
       return root->val+max(0, left);
    }
    
    int maxPathSum(TreeNode *root) {
        if (!root) return 0;
        int result = root->val;
        helper(root, result);
        return result;
    }
};</span>