LeetCode 113. Path Sum II

113. Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]

分析：pathSum函数中只需dfs一下然后返回result数组即可，dfs函数中从root开始寻找到底端sum == root->val的结点，如果满足就将root->val压入path数组中，path数组压入result数组中，然后将当前结点弹出，return。不满足是最后一个结点的则不断深度优先左结点、右结点，同时处理好path数组的压入和弹出～～

/** * 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:    vector<vector<int>> result;    vector<int> path;    vector<vector<int>> pathSum(TreeNode* root, int sum) {        dfs(root, sum);        return result;    }    void dfs(TreeNode* root, int sum) {        if(root == NULL) return ;        if(root->val == sum && root->left == NULL && root->right == NULL) {            path.push_back(root->val);            result.push_back(path);            path.pop_back();            return ;        }        path.push_back(root->val);        dfs(root->left, sum - root->val);        dfs(root->right, sum - root->val);        path.pop_back();    }};