二叉树的镜像

101. Symmetric Tree

判断一棵二叉树是不是镜像对称的。

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:

But the following [1,2,2,null,3,null,3] is not:

Note:

Bonus points if you could solve it both recursively and iteratively.

Solution

explanation: https://leetcode.com/articles/symmetric-tree/

# 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):
    # do it iteratively
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        stk = [(root.left, root.right)]
        while stk:
            n1, n2 = stk.pop()
            if n1 and n2 and n1.val != n2.val:
                return False
            elif n1 and n2:
                stk += [(n1.left, n2.right), (n1.right, n2.left)]
            elif n1 or n2:
                return False
        return True
    # do it recursively
    '''
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root == None:
            return True
        return self.Sym(root.left, root.right)
    
    def Sym(self, left, right):
        if left == None and right == None:
            return True
        if left and right and left.val == right.val:
            return self.Sym(left.left, right.right) and self.Sym(left.right, right.left)
        else:
            return False
    '''

二叉树的镜像

输入一棵二叉树，输出它的镜像。
比如，输入：

输出：

Solution

交换每个节点的左右子节点。

    def MirrorRecursively(self, root):
        if not root:
            return
        if not root.left and not root.right:
            return
        root.left, root.right = root.right, root.left
        if root.left:
            self.MirrorRecursively(root.left)
        if root.right:
            self.MirrorRecursively(root.right)