Leetcode 124: Binary Tree Maximum Path Sum

最新推荐文章于 2024-03-25 21:39:45 发布

weixin_30451709

最新推荐文章于 2024-03-25 21:39:45 发布

阅读量109

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/liangmou/p/7865532.html

本文介绍了一种寻找给定二叉树中最大路径和的算法实现。路径定义为从任意节点开始到任意节点结束的序列，至少包含一个节点，并非必须通过根节点。示例中使用了一个具体二叉树实例进行解释。

Given a 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.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int MaxPathSum(TreeNode root) {
12         var max = new int[1] {Int32.MinValue};
13         DFS(root, max);
14         return max[0];
15     }
16     
17     private int DFS(TreeNode node, int[] max)
18     {
19         if (node == null) return Int32.MinValue;
20         
21         max[0] = Math.Max(max[0], node.val);
22         
23         var left = DFS(node.left, max);
24         var right = DFS(node.right, max);
25 
26         if (left >= 0 && right >= 0)
27         {
28             max[0] = Math.Max(max[0], left + right + node.val);
29             return left >= right ? left + node.val : right + node.val;
30         }
31         else if (left >= 0)
32         {
33             max[0] = Math.Max(max[0], left + node.val);
34             return left + node.val;
35         }
36         else if (right >= 0)
37         {
38             max[0] = Math.Max(max[0], right + node.val);
39             return right + node.val;
40         }
41 
42         return node.val;
43     }
44 }