LeetCode 124. Binary Tree Maximum Path Sum 二叉树中的最大路径和 (C++/Java)

题目：

Given a non-empty 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.

Example 1:

Input: [1,2,3]

       1
      / 
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / 
  9  20
    /  
   15   7

Output: 42

分析：

给定一颗二叉树，求其最大路径，路径就是从任意节点出发，到达任意节点的序列，注意路径至少要有一个节点。

这道题和下面两道题做法相同，都是求二叉树的路径问题，可以先回顾下面两个问题。

LeetCode 543. Diameter of Binary Tree 二叉树的直径 (C++/Java)

LeetCode 687. Longest Univalue Path 最长同值路径 (C++/Java)

那么这道题还是从根节点递归求解，当前结点的最大路径，是当前节点的值加上左右孩子的最大路径，同时和全局的最大值进行比较，更新最大值，而作为返回值时，需要在左右孩子中选取最大值加上当前结点的值同时和0比较，选取最大值，因为节点存在负数，那么路径为负数显然是不对的，所以对于值为负数的子树我们向上层返回0即可。

程序：

C++

/**
 * 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) {
        if(root == nullptr)
            return 0;
        int res = INT_MIN;
        maxPathSum(root, res);
        return res;
    }
private:
    int maxPathSum(TreeNode* root, int& res){
        if(root == nullptr)
            return 0;
        int l = maxPathSum(root->left, res);
        int r = maxPathSum(root->right, res);
        int sum = l + r + root->val;
        res = max(sum, res);
        return max(max(l, r) + root->val, 0);
    }
};

Java

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