regular leetcode

Author: KrisYu

102._binary_tree_level_order_traversal.md

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+###102. Binary Tree Level Order Traversal
+
+题目:
+
+<https://leetcode.com/problems/binary-tree-level-order-traversal/>
+
+
+难度:
+
+Easy
+
+
+我觉得并不easy
+
+两种做法，利用curLevel和nextLevel来记录，然后按层append.
+
+
+```
+class Solution(object):
+    def levelOrder(self, root):
+        """
+        :type root: TreeNode
+        :rtype: List[List[int]]
+        """
+        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
+        
+        return res
+```
+
+
+第二种做法：
+
+
+
+```
+class Solution:
+    # @param root, a tree node
+    # @return a list of lists of integers
+    def preorder(self, root, level, res):
+        if root:
+            if len(res) < level+1: res.append([])
+            res[level].append(root.val)
+            self.preorder(root.left, level+1, res)
+            self.preorder(root.right, level+1, res)
+    def levelOrder(self, root):
+        res=[]
+        self.preorder(root, 0, res)
+        return res
+```
+用递归来记录每一层，需要更加学习，不算easy

103._binary_tree_zigzag_level_order_traversal.md

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+###103. Binary Tree Zigzag Level Order Traversal
+
+题目:
+
+<https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/>
+
+
+难度:
+
+Medium
+
+
+继续用102 的算法作弊
+
+使用102作弊的题目都需要再努力寻求一下别家解法
+
+```
+
+class Solution(object):
+    def zigzagLevelOrder(self, root):
+        """
+        :type root: TreeNode
+        :rtype: List[List[int]]
+        """
+        res = []
+        if root == None: return []
+        
+        curLevel = [root]
+        levelCount = 0
+        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)
+            if levelCount % 2 == 0: 
+                res.append(tmpRes)
+            else : 
+                tmpRes.reverse()
+                res.append(tmpRes)
+            levelCount += 1
+            curLevel = nextLevel
+
+        return res
+```
+
+

107._binary_tree_level_order_traversal_ii.md

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+###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):
+    def levelOrderBottom(self, root):
+        """
+        :type root: TreeNode
+        :rtype: List[List[int]]
+        """
+        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
+```
+
+