From b7b1fba255c94083b9f4ea0e052d5b519461f785 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Fri, 7 Feb 2020 10:52:28 +0800
Subject: [PATCH] update 113

---
 solutions/113. Path Sum II.md | 24 +++++++++++-------------
 1 file changed, 11 insertions(+), 13 deletions(-)

diff --git a/solutions/113. Path Sum II.md b/solutions/113. Path Sum II.md
index 22d48b2..e3bd326 100644
--- a/solutions/113. Path Sum II.md	
+++ b/solutions/113. Path Sum II.md	
@@ -5,6 +5,8 @@
 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。
 
+相当于遍历一遍二叉树，所以时空复杂度均为O(n)
+
 # C++
 ``` C++
 class Solution {
@@ -12,24 +14,20 @@ private:
     vector<int>path;
     vector<vector<int>>res;
     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);
         sum -= root -> val;
-        helper(root -> left, sum);
-        helper(root -> right, sum);
-        path.pop_back(); // 注意离开前要将当前节点移除
+        if((!root->left) && (!root -> right) && 0 == sum) res.push_back(path);
+        else{
+            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:
     vector<vector<int>> pathSum(TreeNode* root, int sum) {
-        path.clear();
-        res.clear();
-        helper(root, sum);
+        path.clear(); res.clear();
+        if(root) helper(root, sum);
         return res;
     }
 };