給定一個非空二叉樹，返回其最大路徑和。

本題中，路徑被定義為一條從樹中任意節點出發，達到任意節點的序列。該路徑至少包含一個節點，且不一定經過根節點。

示例 1:

輸入: [1,2,3]

       1
      / \
     2   3

輸出: 6

示例 2:

輸入: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

輸出: 42

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int ret = INT_MIN;
        onePath(root, ret);
        return ret;
    }
    int onePath(TreeNode *root, int &ret){
        if(root == nullptr)
            return 0;
        int l = onePath(root->left, ret);
        int r = onePath(root->right,ret);
        ret = max(ret,max(0,l)+max(0,r)+root->val);
        return max(0,max(l,r)+root->val);
    }
};