LeetCode 124. 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 must contain at least one node and does not need to go through the root.

For example:
Given the below binary tree,

   1  / \ 2   3

Return 6.

求二叉树任意起止点构成的路径的值最大和

虽然是任意起止点，但此路径一定会有根节点，因此对于该根节点：
max = root.val+maxPathSum(左子树)(>0才加)+maxPathSum(右子树)(>0才加);

但对于每次递归的返回，则应该是
return max(root.val,root.val+maxPathSum(左子树),root.val+maxPathSum(右子树));

 public static int max;            public static int maxPathSum(TreeNode root) {                if(root==null)                    return 0;                max = Integer.MIN_VALUE;                maxPathSum2(root);                return max;            }            public static int maxPathSum2(TreeNode root) {                if(root==null)                    return 0;                int temp1 = maxPathSum2(root.left);                int temp2 = maxPathSum2(root.right);                int temp = root.val;                if(temp1>0)temp+=temp1;                if(temp2>0)temp += temp2;                max = temp>max?temp:max;                return max(root.val,root.val+temp1,root.val+temp2);            }            public static int max(int i,int j,int k){                return i > (j>k?j:k) ? i : (j>k?j:k);            }