101. 对称二叉树

我的思路为中序遍历和逆中序遍历的结果是相同的。这种思路是错的，原因在于
[1,2,2,2,null,2]这种情况下回出现错误。

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        
        queue1 = []
        queue2 = []
        self.Inorder(root,queue1)
        self.inverseInorder(root, queue2)
        if queue1 == queue2:
            return True
        else:
            return False
        
    def Inorder(self, root,queue):
        if root is None:
            return 
        self.Inorder(root.left,queue)
        queue.append(root.val)
        self.Inorder(root.right,queue)
        
    def inverseInorder(self, root,queue):
        if root is None:
            return 
        self.inverseInorder(root.right,queue)
        queue.append(root.val)
        self.inverseInorder(root.left,queue)

随后看了官方解答，进行了递归算法的实现。

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        return self.isMirror(root,root)
    def isMirror(self, rootleft, rootright):
        if rootleft == None and rootright == None:
            return True
        if rootleft == None or rootright == None:
            return False
        if rootleft.val != rootright.val:
            return False
        
        return self.isMirror(rootleft.left, rootright.right) and self.isMirror(rootleft.right, rootright.left)

因为通过递归的方式可以实现，自然想到通过栈的方式（其实在这道题中用队列也可以实现）。

# 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
        stack = [root.left,root.right]
        
        while stack:
            t1 = stack.pop()
            t2 = stack.pop()
            if (not t1 and not t2):
                continue
            if(not t1 or not t2):
                return False
            if (t1.val != t2.val):
                return False
            stack.append(t1.left)
            stack.append(t2.right)
            stack.append(t1.right)
            stack.append(t2.left)
        return True