From be65903f540a913bbda74ab32a9644da3d024364 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Mon, 10 Jun 2019 23:51:58 +0800
Subject: [PATCH] add 114

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+# [114. Flatten Binary Tree to Linked List](https://leetcode.com/problems/flatten-binary-tree-to-linked-list/)
+
+# 思路
+要求把二叉树展开成链表，根据例子可以看到是先序遍历的顺序。
+## 思路一
+最容易想到的就是先序遍历一遍树。我们可以用一个全局的指针`list_last`记录当前链表的最后一个节点，每遍历到一个节点就将其接到`list_last`后面（右指针）。
+
+## 思路二
+此题也可以直接递归来做。即如果root的左右子树都已经是被展开好的，那么只需要将左子树接在root的右边，再将右子树接在原来左子树的最下边就可以了，如下例。
+```
+  1               1
+ / \               \
+2   5     ===>      2            
+ \   \               \
+  3   6               3
+   \                   \
+    4                   4
+                         \
+                          5
+                           \
+                            6
+```
+
+# C++
+## 思路一
+``` C++
+class Solution {
+private:
+    TreeNode *list_last = NULL;
+    
+    void helper(TreeNode *root){ // 先序遍历
+        if(root == NULL) return;
+        if(list_last != NULL)  // 不是第一个
+            list_last -> right = root;
+        
+        list_last = root;
+        
+        // 由于后面要将左右指针置空,所以这里先记录一下
+        TreeNode *left = root -> left; 
+        TreeNode *right = root -> right;
+        root -> left = NULL;
+        root -> right = NULL;
+        
+        helper(left);
+        helper(right);
+        
+    }
+public:
+    void flatten(TreeNode* root) {
+        if(root == NULL) return;
+        list_last = NULL;
+        helper(root);
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+public:
+    void flatten(TreeNode* root) {
+        if(root == NULL) return; // 递归出口
+        flatten(root -> left);
+        flatten(root -> right);
+        
+        TreeNode *right = root -> right; // 先备份右子树
+        root -> right = root -> left; // 将左子树放在右边
+        root -> left = NULL;
+        
+        TreeNode *tmp = root;
+        // 一路下降到最下面
+        while(tmp -> right) tmp = tmp -> right;
+        tmp -> right = right;
+    }
+};
+```
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