LeetCode113. Path Sum II(Medium)

原题：https://leetcode.com/problems/path-sum-ii/description/

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题目描述：
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]]

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题解
和上一题解题思路一样。每次递归时，都将新的节点的值存入当前vector< int >中。如果当前节点是叶子节点且路径和等于sum，那就将当前vector< int >存入一个vector< vector< int >>变量中。因为每次递归vector< int >变量使用的都不是引用，所以每次调用递归时都会在函数中生成一个传递参数的副本，虽然浪费空间但是正好适合递归的做法。

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代码：

/** * 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:    void findPath(TreeNode* root, int sum, vector<int> v, vector<vector<int>>& vs) {        if (!root) return;        v.push_back(root->val);        if (root->val == sum&&!root->left&&!root->right) vs.push_back(v);        sum -= root->val;        findPath(root->left, sum, v, vs);        findPath(root->right, sum, v, vs);        vector<int>().swap(v);    }    vector<vector<int>> pathSum(TreeNode* root, int sum) {        vector<vector<int>> vs;        findPath(root, sum, vector<int>(), vs);        return vs;    }};

114 / 114 test cases passed.
Status: Accepted
Runtime: 12 ms`