Update 107._binary_tree_level_order_traversal_ii.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/107._binary_tree_level_order_traversal_ii.md

@@ -1,48 +1,61 @@
-###107. Binary Tree Level Order Traversal II
-
-题目:
-
-<https://leetcode.com/problems/binary-tree-level-order-traversal-ii/>
-
-
-难度:
-
-Easy
-
-
-用102 的算法作弊
-
-
-```
-# 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):
+# 107. Binary Tree Level Order Traversal II
+
+**<font color=red>难度: Easy</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/binary-tree-level-order-traversal-ii/description/
+
+> 内容描述
+
+```
+Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
+
+For example:
+Given binary tree [3,9,20,null,null,15,7],
+    3
+   / \
+  9  20
+    /  \
+   15   7
+return its bottom-up level order traversal as:
+[
+  [15,7],
+  [9,20],
+  [3]
+]
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(N)******
+
+
+用102 的算法作弊
+
+
+```python
+class Solution:
     def levelOrderBottom(self, root):
         """
         :type root: TreeNode
         :rtype: List[List[int]]
         """
+        def dfs(node, level, res):
+            if not node:
+                return
+            if len(res) < level:
+                res.append([])
+            res[level-1].append(node.val)
+            dfs(node.left, level+1, res)
+            dfs(node.right, level+1, res)  
+                
         res = []
-        
-        if root == None: return []
-        
-        curLevel = [root]
-        while curLevel:
-            nextLevel = []
-            tmpRes = []
-            for node in curLevel:
-                tmpRes.append(node.val)
-                if node.left: nextLevel.append(node.left)
-                if node.right: nextLevel.append(node.right)
-            res.append(tmpRes)
-            curLevel = nextLevel
-        res.reverse()
-        return res
-```
-
-
+        dfs(root, 1, res)
+        return res[::-1]
+```
+
+