leetCode练习（124）

题目：Binary Tree Maximum Path Sum

难度：hard

问题描述：

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 does not need to go through the root.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

解题思路：

通过递归，求解每个经过最高层节点n的最大值。经过‘1’节点的最大值=1+1.left的单边最大值+1.right的单边最大值。也就是说，对于节点n，其返回值为n.val加上其中一条边或者0条边的值。同时设置全局变量保存每各经过的最高层节点路径和中的最大者。

代码如下：

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