#114- flatten-binary-tree-to-linked-list

daily/2019-06-14.md

@@ -0,0 +1,158 @@
+## 每日一题 - flatten-binary-tree-to-linked-list
+
+### 信息卡片
+
+- 时间：2019-06-14
+- 题目链接：https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list/
+- tag：`tree`  `Recursion `
+
+### 题目描述
+
+```
+Given a binary tree, flatten it to a linked list in-place.
+
+For example, given the following tree:
+
+    1
+   / \
+  2   5
+ / \   \
+3   4   6
+The flattened tree should look like:
+
+1
+ \
+  2
+   \
+    3
+     \
+      4
+       \
+        5
+         \
+          6
+```
+
+### 参考答案
+
+#### 方法1 先序遍历
+
+如果仔细观察输入输出的话会发现，其实输出其实就是输入的先序遍历结果而已。
+因此一种做法就是我们对其进行先序遍历，
+
+然后将先序遍历的结果构造成没有左子树的二叉树即可
+
+Time complexity : O(n)
+Space complexity : O(n)
+
+参考代码
+```javascript
+/*
+ * @lc app=leetcode id=114 lang=javascript
+ *
+ * [114] Flatten Binary Tree to Linked List
+ */
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+function preorderTraversal(root) {
+  if (!root) return [];
+
+  return [root]
+    .concat(preorderTraversal(root.left))
+    .concat(preorderTraversal(root.right));
+}
+/**
+ * @param {TreeNode} root
+ * @return {void} Do not return anything, modify root in-place instead.
+ */
+var flatten = function(root) {
+    if (root === null) return root;
+    const res = preorderTraversal(root);
+
+    let curPos = 0;
+    let curNode = res[0];
+
+    while(curNode = res[curPos]) {
+        curNode.left = null;
+        curNode.right = res[++curPos];
+    }
+};
+```
+
+
+
+### 其他优秀解答
+
+#### 方法2  先序遍历优化（递归遍历和非递归遍历）
+算法描述
+
+ 把一颗二叉树变成单链表  flatten(root)
+
+- 递归遍历把一棵树左子树变成单链表 a
+- 递归遍历把一棵树右子树变成单链表 b
+- 用链表a最后一个元素拼接链表b（递归子问题）
+
+参考代码
+
+- 递归
+```c++
+
+   void flatten(TreeNode* root) {
+       if (root == NULL) return ;
+
+       flatten(root->left);
+       flatten(root->right);
+
+       //递归子问题
+       TreeNode *tmp = root->right;
+       root->right = root->left; 
+       root->left = NULL;
+
+       while (root->right)
+       {
+         root = root->right; 
+       };
+
+        root->right = tmp; 
+   }
+  };
+```
+
+-  非递归
+```c++
+
+ void flatten(TreeNode* root) {
+     if (root == NULL) {
+          return ;
+     }
+     stack<TreeNode*> result;
+     result.push(root);
+
+     while (!result.empty()){
+         TreeNode* cur=result.top();
+         result.pop();
+
+         if (cur->right)
+         {
+            result.push(cur->right);//先顺非递归遍历
+         }
+
+         if (cur->left)
+         {
+            result.push(cur->left);//先顺非递归遍历
+         }
+         //递归子问题
+         if (!result.empty()) 
+         {
+             cur->right=result.top();
+         }
+         cur->left=NULL;
+     }
+
+
+```

daily/README.md

@@ -47,3 +47,9 @@ tag: `Tree`
 tag: `Probability Theory` 
 
 时间: 2019-06-13
+
+### [114.flatten-binary-tree-to-linked-list](./2019-06-14.md)
+
+tag: `tree` 
+
+时间: 2019-06-14