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[lintcode] Binary Tree Maximum Path Sum II

时间:2016-08-15 17:13:54      阅读:161      评论:0      收藏:0      [点我收藏+]

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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

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