124 Binary Tree Maximum Path Sum

最新推荐文章于 2024-03-25 21:39:45 发布

dongbeier

最新推荐文章于 2024-03-25 21:39:45 发布

阅读量118

点赞数

CC 4.0 BY-SA版权

分类专栏： MicroSoft LeetCode

本文链接：https://blog.youkuaiyun.com/dongbeier/article/details/80345404

LeetCode 同时被 2 个专栏收录

146 篇文章

订阅专栏

MicroSoft

85 篇文章

订阅专栏

本文介绍了一种使用深度优先搜索(DFS)算法解决二叉树最大路径和问题的方法。通过递归计算左子树和右子树的最大贡献值，并结合当前节点值，找出整棵树中的最大路径和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

用Binary
Tree最常用的dfs来进行遍历。先算出左右子树的结果L和R，如果L大于0，那么对后续结果是有利的，我们加上L，如果R大于0，对后续结果也是有利的，继续加上R

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    private int maxSum;
    
    public int maxPathSum(TreeNode root) {
        maxSum = Integer.MIN_VALUE;
        dfs(root);
        return maxSum;
    }
    
    private int dfs(TreeNode root) {
        if (root == null)
            return 0;
        int left = dfs(root.left);
        int right = dfs(root.right);
        int sum = root.val;
        if (left > 0)
            sum += left;
        if (right > 0)
            sum += right;
        maxSum = Math.max(maxSum, sum);
        return Math.max(left, right) > 0 ? Math.max(left, right) +
            root.val : root.val;
    }
}