LeetCode101：Symmetric Tree

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:

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LeetCode：链接

该题的做法和LeetCode100：Same Tree非常类似。

第一种：递归

# 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 root == None:
            return True
        return self.isMirror(root.left, root.right)

    def isMirror(self, left, right):
        if left == None and right == None:
            return True
        if left == None or right == None:
            return False
        if left.val != right.val:
            return False
        return self.isMirror(left.left, right.right) and self.isMirror(left.right, right.left)

第二种：非递归(DFS)

# 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
        stack = [(root.left, root.right)]
        while stack:
            left, right = stack.pop()
            if left == None and right == None:
                continue
            elif left == None or right == None:
                return False
            elif left.val == right.val:
                stack.append((left.left, right.right))
                stack.append((left.right, right.left))
            else:
                return False
        return True