From 568de5d114cfe672375b872361cbf0f6b6b9664f Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Fri, 31 May 2019 23:38:32 +0800
Subject: [PATCH] Create 113. Path Sum II.md

---
 solutions/113. Path Sum II.md | 36 +++++++++++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)
 create mode 100644 solutions/113. Path Sum II.md

diff --git a/solutions/113. Path Sum II.md b/solutions/113. Path Sum II.md
new file mode 100644
index 0000000..22d48b2
--- /dev/null
+++ b/solutions/113. Path Sum II.md	
@@ -0,0 +1,36 @@
+# [113. Path Sum II](https://leetcode.com/problems/path-sum-ii/)
+# 思路
+给定一棵二叉树，求所有根到叶子路径中和为sum的路径。       
+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
+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。
+
+# C++
+``` C++
+class Solution {
+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
+    }
+public:
+    vector<vector<int>> pathSum(TreeNode* root, int sum) {
+        path.clear();
+        res.clear();
+        helper(root, sum);
+        return res;
+    }
+};
+```