leetcode 437. Path Sum III 深度优先遍历DFS

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1

Return 3. The paths that sum to 8 are:

1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11

题意很简单，就是DFS深度优先遍历求解

建议和leetcode 112. Path Sum DFS深度优先遍历 和leetcode 113. Path Sum II DFS深度优先遍历 一起学习

还有leetcode 124. Binary Tree Maximum Path Sum 最大路径和 + DFS深度优先搜索 和 leetcode 687. Longest Univalue Path 深度优先遍历DFS 一起学习

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>using namespace std;/*struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {}};*/class Solution {public:    int pathSum(TreeNode* root, int sum)     {        if (root == NULL)             return 0;        return sumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);    }private:    int sumUp(TreeNode* root, int pre, int& sum)    {        if (root==NULL)            return 0;        int current = pre + root->val;        return (current == sum) + sumUp(root->left, current, sum) + sumUp(root->right, current, sum);    }};