sovle problem Flatten Binary Tree To Linked List

Author: P-ppc

README.md

@@ -237,6 +237,7 @@ All solutions will be accepted!
 |105|[Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/)|[java/py/js](./algorithms/ConstructBinaryTreeFromPreorderAndInorderTraversal)|Medium|
 |654|[Maximum Binary Tree](https://leetcode-cn.com/problems/maximum-binary-tree/description/)|[java/py/js](./algorithms/MaximumBinaryTree)|Medium|
 |208|[Implement Trie Prefix Tree](https://leetcode-cn.com/problems/implement-trie-prefix-tree/description/)|[java/py/js](./algorithms/ImplementTriePrefixTree)|Medium|
+|114|[Flatten Binary Tree To Linked List](https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list/description/)|[java/py/js](./algorithms/FlattenBinaryTreeToLinkedList)|Medium|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/FlattenBinaryTreeToLinkedList/README.md

@@ -0,0 +1,2 @@
+# Flatten Binary Tree To Linked List
+This problem is easy to solve by DFS

algorithms/FlattenBinaryTreeToLinkedList/Solution.java

@@ -0,0 +1,28 @@
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public void flatten(TreeNode root) {
+        LinkedList<TreeNode> rights = new LinkedList<TreeNode>();
+        
+        while (root != null) {
+            if (root.left != null && root.right != null) {
+                rights.addFirst(root.right);
+            }
+            
+            if (root.left != null) {
+                root.right = root.left;
+                root.left = null;
+            } else if (root.right == null && rights.size() > 0) {
+                root.right = rights.removeFirst();
+            }
+            root = root.right;
+        }
+    }
+}

algorithms/FlattenBinaryTreeToLinkedList/solution.js

@@ -0,0 +1,27 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {void} Do not return anything, modify root in-place instead.
+ */
+var flatten = function(root) {
+    let rights = []
+    
+    while (root) {
+        if (root.left && root.right) {
+            rights.push(root.right)
+        }
+        if (root.left) {
+            root.right = root.left
+            root.left = null
+        } else if (!root.right && rights.length > 0) {
+            root.right = rights.pop()
+        }
+        root = root.right
+    }
+};

algorithms/FlattenBinaryTreeToLinkedList/solution.py

@@ -0,0 +1,26 @@
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution(object):
+    def flatten(self, root):
+        """
+        :type root: TreeNode
+        :rtype: void Do not return anything, modify root in-place instead.
+        """
+        rights = []
+        
+        node = root
+        while node:
+            if node.left and node.right:
+                rights.append(node.right)
+                
+            if node.left:
+                node.right = node.left
+                node.left = None
+            elif len(rights) > 0 and not node.right:
+                node.right = rights.pop()
+            node = node.right