LeetCode：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]]

解题分析：

此题和Path Sum不太一样，这道题是要把所有的路径都求出来，而不只是返回一个找到与否的bool值。

要把路径记录下来，当一个结点的左右结点访问完后返回时，应当把该结点剔除掉，也就是说存储路径的容器应当具备栈后进先出的特点，在c++中用vector也可以实现，另外还需要一个二维的向量来保存所有路径。

代码如下：

class Solution {public:    vector<vector<int>> pathSum(TreeNode* root, int sum) {        vector<vector<int> > paths;        vector<int> path;        findPaths(root, sum, path, paths);        return paths;      }private:    void findPaths(TreeNode* node, int sum, vector<int>& path, vector<vector<int> >& paths) {        if (!node) return;        path.push_back(node -> val);        if (!(node -> left) && !(node -> right) && sum == node -> val)            paths.push_back(path);        findPaths(node -> left, sum - node -> val, path, paths);        findPaths(node -> right, sum - node -> val, path, paths);        path.pop_back();    }};