113. Path Sum II

问题描述：

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：该问题必须遍历每条路径，一次判断是否该路径使我们需要的路径

代码：

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int> > allPaths;//返回的路径
        vector<int> path;
        findPaths(root, sum, path, allPaths);
        return allPaths;  
    }
private:
    void findPaths(TreeNode* node, int sum, vector<int>& path, vector<vector<int> >& paths) {
        if (!node) return;//节点为空时
        path.push_back(node -> val);//不为空时，将当前节点加入path中
        if (!(node -> left) && !(node -> right) && sum == node -> val)//一条路劲结束时判断是否为需要的路径
            paths.push_back(path);
        if(node -> left  && node->val<=sum)//还有子节点时，若当前路径的和必须小于会等于给出的和
            findPaths(node -> left, sum - node -> val, path, paths);
       if(node -> right  && node->val<=sum)
            findPaths(node -> right, sum - node -> val, path, paths);
        path.pop_back();
    }
};