LeetCode Binary Tree Maximum Path Sum

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.

题目：求二叉树中和最大的路径。

思路：可以这么想：可能的路径都是要经过一个类根节点的，那么结果无非就是（根节点+左路径的最大值+右路径的最大值）中的最大值，这是结果的筛选，但是我们递归返回值的时候并不是上面的值，而应该是路径的最大值也就是左子树中的路径、右子树的路径或者自己，然后就能递归求解了。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int ans;
	
	public int cal(TreeNode root) {
		if (root == null) return 0;
		int left = cal(root.left);
		int right = cal(root.right);
		int val = root.val;
		int newRootVal = Math.max(val, Math.max(left+val, right+val));
		if (left > 0) val += left;
		if (right > 0) val += right;
		ans = Math.max(ans, val);
		return newRootVal;
	}
	
    public int maxPathSum(TreeNode root) {
    	ans = 0;
    	if (root == null) return 0;
    	ans = root.val;
    	cal(root);
        return ans; 
    }
}