2019-05-31 15:38:32 +00:00
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# [113. Path Sum II](https://leetcode.com/problems/path-sum-ii/)
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# 思路
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给定一棵二叉树，求所有根到叶子路径中和为sum的路径。
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考虑用DFS，遍历的时候用一维数组path记录当前经过的路径，每遍历一个节点就将其加入path，从这个结点返回时，需要把该节点从path中移除。如果当前节点是叶子节点而且
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满足和为sum，即可将path加入到结果二维数组res中。
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注：实现的时候我们可以将path和res设置成全局的，免得传参。
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2020-02-07 02:52:28 +00:00
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相当于遍历一遍二叉树，所以时空复杂度均为O(n)
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2019-05-31 15:38:32 +00:00
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# C++
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``` C++
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class Solution {
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private:
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vector<int>path;
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vector<vector<int>>res;
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void helper(TreeNode *root, int sum){
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path.push_back(root -> val);
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sum -= root -> val;
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2020-02-07 02:52:28 +00:00
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if((!root->left) && (!root -> right) && 0 == sum) res.push_back(path);
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else{
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if(root -> left) helper(root -> left, sum);
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if(root -> right) helper(root -> right, sum);
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}
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// 返回父节点之前要删除路径上的当前节点
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path.pop_back();
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2019-05-31 15:38:32 +00:00
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}
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public:
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vector<vector<int>> pathSum(TreeNode* root, int sum) {
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2020-02-07 02:52:28 +00:00
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path.clear(); res.clear();
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if(root) helper(root, sum);
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2019-05-31 15:38:32 +00:00
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return res;
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}
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};
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```
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