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update 113

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ShusenTang 2020-02-07 10:52:28 +08:00
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commit b7b1fba255
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solutions/113. Path Sum II.md
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@ -5,6 +5,8 @@
满足和为sum，即可将path加入到结果二维数组res中。 满足和为sum，即可将path加入到结果二维数组res中。
注：实现的时候我们可以将path和res设置成全局的，免得传参。 注：实现的时候我们可以将path和res设置成全局的，免得传参。
相当于遍历一遍二叉树，所以时空复杂度均为O(n)
# C++ # C++
``` C++ ``` C++
class Solution { class Solution {
@ -12,24 +14,20 @@ private:
vector<int>path; vector<int>path;
vector<vector<int>>res; vector<vector<int>>res;
void helper(TreeNode *root, int sum){ void helper(TreeNode *root, int sum){
if(root == NULL) return;
if(root -> val == sum && (!root->left) && (!root -> right)){ // 满足条件
path.push_back(root -> val);
res.push_back(path);
path.pop_back();
return;
}
path.push_back(root -> val); path.push_back(root -> val);
sum -= root -> val; sum -= root -> val;
helper(root -> left, sum); if((!root->left) && (!root -> right) && 0 == sum) res.push_back(path);
helper(root -> right, sum); else{
path.pop_back(); // 注意离开前要将当前节点移除 if(root -> left) helper(root -> left, sum);
if(root -> right) helper(root -> right, sum);
}
// 返回父节点之前要删除路径上的当前节点
path.pop_back();
} }
public: public:
vector<vector<int>> pathSum(TreeNode* root, int sum) { vector<vector<int>> pathSum(TreeNode* root, int sum) {
path.clear(); path.clear(); res.clear();
res.clear(); if(root) helper(root, sum);
helper(root, sum);
return res; return res;
} }
}; };