需要判断输入的结点为空或者输入的结点没有子树的情况。

'''
操作给定的二叉树，将其变换为源二叉树的镜像。
'''

# -*- coding:utf-8 -*-
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
class Solution:
    # 递归实现
    def Mirror(self, root):
        if root == None:
            return
        if root.left == None and root.right == None:
            return root

        pTemp = root.left
        root.left = root.right
        root.right = pTemp

        self.Mirror(root.left)
        self.Mirror(root.right)

    # 非递归实现
    def Mirror2(self, root):
        if root == None:
            return
        stackNode = []
        stackNode.append(root)
        while len(stackNode) > 0:
            nodeNum = len(stackNode) - 1
            tree = stackNode[nodeNum]
            stackNode.pop()
            nodeNum -= 1
            if tree.left != None or tree.right != None:
                tree.left, tree.right = tree.right, tree.left
            if tree.left:
                stackNode.append(tree.left)
                nodeNum += 1
            if tree.right:
                stackNode.append(tree.right)
                nodeNum += 1
    # 非递归实现
    def MirrorNoRecursion(self, root):
        if root == None:
            return
        nodeQue = [root]
        while len(nodeQue) > 0:
            curLevel, count = len(nodeQue), 0
            while count < curLevel:
                count += 1
                pRoot = nodeQue.pop(0)
                pRoot.left, pRoot.right = pRoot.right, pRoot.left
                if pRoot.left:
                    nodeQue.append(pRoot.left)
                if pRoot.right:
                    nodeQue.append(pRoot.right)

pNode1 = TreeNode(8)
pNode2 = TreeNode(6)
pNode3 = TreeNode(10)
pNode4 = TreeNode(5)
pNode5 = TreeNode(7)
pNode6 = TreeNode(9)
pNode7 = TreeNode(11)

pNode1.left = pNode2
pNode1.right = pNode3
pNode2.left = pNode4
pNode2.right = pNode5
pNode3.left = pNode6
pNode3.right = pNode7

S = Solution()
S.Mirror2(pNode1)
print(pNode1.right.left.val)