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Given a binary tree, find the maximum path sum from root.
The path may end at any node in the tree and contain at least one node in it.
给一棵二叉树,找出从根节点出发的路径中,和最大的一条。
这条路径可以在任何二叉树中的节点结束,但是必须包含至少一个点(也就是根了)。
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root the root of binary tree. * @return an integer */ public int maxPathSum2(TreeNode root) { if (root == null) { return Integer.MIN_VALUE; } int left = maxPathSum2(root.left); int right = maxPathSum2(root.right); return root.val + Math.max(0, Math.max(left, right)); } }
[lintcode] Binary Tree Maximum Path Sum II
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原文地址:http://www.cnblogs.com/iwangzheng/p/5773598.html