leetcode 124. Binary Tree Maximum Path Sum 最大路径和 + DFS深度优先搜索

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.

这道题求的是通过父子关系连接起来的最大的路径和，路径的起始结点不一定是叶子节点。

我看了这道题好久才明白，那么我们可以利用DFS深度优先搜索来得到左右结点的最大sum，然后更新最大的路径sum。

建议和leetcode 437. Path Sum III 深度优先遍历DFS 和leetcode 687. Longest Univalue Path 深度优先遍历DFS 一起学习

代码如下：

/*class TreeNode {      int val;      TreeNode left;      TreeNode right;      TreeNode(int x) { val = x; }}*//* * 函数递归返回的是某一路径的最大sum * 那么对于某一个结点，更新相关详细即可 * 如果只是一个节点,那么当然就是这个节点的值了. *  * */public class Solution{    int MaxPathSum = Integer.MIN_VALUE;    public int maxPathSum(TreeNode root)     {        maxPathSumByRecursion(root);        return MaxPathSum;    }    public int maxPathSumByRecursion(TreeNode root)     {        if(root==null)            return 0;        else        {            int leftMax = maxPathSumByRecursion(root.left);            int rightMax = maxPathSumByRecursion(root.right);            int tmp = root.val;            if(leftMax>0)                tmp += leftMax;            if(rightMax>0)                tmp += rightMax;            //这里更新最大路径值            MaxPathSum = Math.max(MaxPathSum, tmp);            //这里返回的是当前左孩子或者右孩子的最大的sum，并不是某一条路径的最大sum            return Math.max(root.val, Math.max(leftMax + root.val, rightMax + root.val));        }    }}

下面手C++的做法，就是DFS深度优先遍历的做法，这里要注意两个问题：
我们是用DFS递归来计算某一个路径的sum，然后在递归计算中间计算比较最大的路径值

代码如下：

#include <iostream>#include <vector>#include <algorithm>#include <climits>using namespace std;/*struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {}};*/class Solution {public:    int maxSum = numeric_limits<int>::min();    int maxPathSum(TreeNode* root)     {        if (root == NULL)            return 0;        byDFS(root);        return maxSum;    }    int byDFS(TreeNode* root)    {        if (root == NULL)            return 0;        else        {            int left = byDFS(root->left);            int right = byDFS(root->right);            int tmp = root->val;            if (left > 0)                tmp += left;            if (right > 0)                tmp += right;            maxSum = max(maxSum, tmp);            return max(root->val,max(left+root->val,right + root->val));        }    }};