lintcode475- Binary Tree Maximum Path Sum II- medium

/**
 * 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
     */
     
    private int maxSum = Integer.MIN_VALUE;
    
    public int maxPathSum2(TreeNode root) {
        // write your code here
        helper(root, 0);
        return maxSum;
    }
    
    // 尝试加此时节点。
    // 如果根固定，儿子末端不固定，那要从上往下走(在参数列表里记录)，不好从下往上递归(在返回值里记录)
    private void helper(TreeNode head, int sum) {
       if (head == null) {
           return;
       } 
       sum += head.val;
       if (sum > maxSum) {
           maxSum = sum;
       }
       
       helper(head.left, sum);
       helper(head.right, sum);
       return;
    }
}

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));
    }
}