solve problem Construct String From Binary Tree

Author: P-ppc

README.md

@@ -83,6 +83,7 @@ All solutions will be accepted!
 |762|[Prime Number Of Set Bits In Binary Representation](https://leetcode-cn.com/problems/prime-number-of-set-bits-in-binary-representation/description/)|[java/py/js](./algorithms/PrimeNumberOfSetBitsInBinaryRepresentation)|Easy|
 |521|[Longest Uncommon Subsequence I](https://leetcode-cn.com/problems/longest-uncommon-subsequence-i/description/)|[java/py/js](./algorithms/LongestUncommonSubsequenceI)|Easy|
 |812|[Largest Triangle Area](https://leetcode-cn.com/problems/largest-triangle-area/description/)|[java/py/js](./algorithms/LargestTriangleArea)|Easy|
+|606|[Construct String From Binary Tree](https://leetcode-cn.com/problems/construct-string-from-binary-tree/description/)|[java/py/js](./algorithms/ConstructStringFromBinaryTree)|Easy|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/ConstructStringFromBinaryTree/README.md

@@ -0,0 +1,2 @@
+# Construct String From Binary Tree
+This problem is easy to solve by recursion

algorithms/ConstructStringFromBinaryTree/Solution.java

@@ -0,0 +1,22 @@
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public String tree2str(TreeNode t) {
+        if (t == null) {
+            return "";
+        } else if (t.left == null && t.right == null) {
+            return String.valueOf(t.val);
+        } else if (t.right == null) {
+            return t.val + "(" + tree2str(t.left) + ")";
+        } else {
+            return t.val + "(" + tree2str(t.left) + ")(" + tree2str(t.right) + ")";
+        }
+    }
+}

algorithms/ConstructStringFromBinaryTree/solution.js

@@ -0,0 +1,22 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {TreeNode} t
+ * @return {string}
+ */
+var tree2str = function(t) {
+    if (t === null) {
+        return ''
+    } else if (t.left === null && t.right === null) {
+        return `${t.val}`
+    } else if (t.right === null) {
+        return `${t.val}(${tree2str(t.left)})`
+    } else {
+         return `${t.val}(${tree2str(t.left)})(${tree2str(t.right)})`
+    }
+};

algorithms/ConstructStringFromBinaryTree/solution.py

@@ -0,0 +1,21 @@
+# 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 tree2str(self, t):
+        """
+        :type t: TreeNode
+        :rtype: str
+        """
+        if not t:
+            return ''
+        elif not t.left and not t.right:
+            return str(t.val)
+        elif not t.right:
+            return str(t.val) + '(' + self.tree2str(t.left) + ')'
+        else:
+            return str(t.val) + '(' + self.tree2str(t.left) + ')' + '(' + self.tree2str(t.right) + ')'