[LeetCode]Binary Tree Maximum Path Sum

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1      / \     2   3

Return 6.

maxPath记录最大的路径，计算方式是

    int sum = root.val;    if(left>0) sum += left;    if(right>0) sum += right;    maxPath = Math.max(sum, maxPath);

然后深搜每个节点，该节点返回的是这个节点的值加上较大的左子树或者较大的右子树

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {int maxPath = Integer.MIN_VALUE;    public int maxPathSum(TreeNode root) {        dfs(root);        return maxPath;    }        private int dfs(TreeNode root){    if(root==null) return 0;    int left = dfs(root.left);    int right =  dfs(root.right);    int sum = root.val;    if(left>0) sum += left;    if(right>0) sum += right;    maxPath = Math.max(sum, maxPath);    return Math.max(left, right)>0?root.val+Math.max(left, right):root.val;    }}