symmetric tree easy

Author: weiy

Tree/SymmetricTree.py

@@ -0,0 +1,91 @@
+"""
+Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
+
+For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
+
+    1
+   / \
+  2   2
+ / \ / \
+3  4 4  3
+But the following [1,2,2,null,3,null,3] is not:
+    1
+   / \
+  2   2
+   \   \
+   3    3
+Note:
+Bonus points if you could solve it both recursively and iteratively.
+
+
+判断左子树是否与右子树互为镜像。
+
+思路：
+遍历左子树，把结果放入队列中。
+按照相反的左右遍历右子树，同时让压出队列顶的一项作对比。
+不匹配或队列中没有足够的数据都可判为False.
+
+效率 O(n)
+beat 100%.
+
+测试地址：
+https://leetcode.com/problems/symmetric-tree/description/
+
+"""
+# 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 isSymmetric(self, root):
+        """
+        :type root: TreeNode
+        :rtype: bool
+        """
+        if not root:
+            return True
+        
+        left = root.left
+        right = root.right
+        
+        left_node = [left]
+        right_node = [right]
+        
+        left_node_value = []
+        
+        while left_node:
+            # if left_node:
+            _left = left_node.pop()
+            if _left:
+                left_node_value.append(_left.val)        
+            else:
+                left_node_value.append(None)
+
+            if _left:
+                left_node.append(_left.right)
+                left_node.append(_left.left)
+        
+        left_node_value.reverse()
+        
+        while right_node:
+            _right = right_node.pop()
+            
+            if left_node_value:
+                left_value = left_node_value.pop()
+            else:
+                return False
+            
+            if _right:
+                if left_value != _right.val:
+                    return False
+            else:
+                if left_value != None:
+                    return False
+
+            if _right:
+                right_node.append(_right.left)
+                right_node.append(_right.right)
+        return True