## Description

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.

## Solutions

### 1. Recurrence

先判断左右子节点是否相等，再判断左子节点的左子节点与右子节点的右子节点，还有左子节点的右子节点和右子节点的左子节点是否一样。

    0
/ \
l   r
/ \ / \
l  r l  r

# Time: O(n)
# Space: O(1)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
return self.is_mirror(root.left, root.right)

def is_mirror(self, left, right):
if not left or not right:
return left == right
if left.val != right.val:
return False
return self.is_mirror(left.left, right.right) and \
self.is_mirror(left.right, right.left)

# 195/195 cases passed (32 ms)
# Your runtime beats 69.89 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.9 MB)


### 2. Iterative-Stack

注意用两个栈！

# Time: O(n)
# Space: O(n)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
left_stack, right_stack = [root.left], [root.right]
while len(left_stack) > 0 and len(right_stack) > 0:
left = left_stack.pop()
right = right_stack.pop()
if not left and not right:
continue
elif not left or not right:
return False
if left.val != right.val:
return False
left_stack.append(left.left)
right_stack.append(right.right)
left_stack.append(left.right)
right_stack.append(right.left)
return len(left_stack) == 0 and len(right_stack) == 0

# 195/195 cases passed (32 ms)
# Your runtime beats 69.89 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.9 MB)


### 3. Iterative-Queue

# Time: O(n)
# Space: O(n)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
left_queue, right_queue = collections.deque([root.left]), collections.deque([root.right])
while len(left_queue) > 0 and len(right_queue) > 0:
left = left_queue.popleft()
right = right_queue.popleft()
if not left and not right:
continue
elif not left or not right:
return False
if left.val != right.val:
return False
left_queue.append(left.left)
right_queue.append(right.right)
left_queue.append(left.right)
right_queue.append(right.left)
return len(left_queue) == 0 and len(right_queue) == 0

# 195/195 cases passed (36 ms)
# Your runtime beats 33.94 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.9 MB)