leetcood学习笔记-101-对称二叉树

题目描述：

方法一：递归：

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        if not root: return True
        def Tree(p, q):
            if not p and not q: return True
            if p and q and p.val == q.val :
                return Tree(p.left, q.right) and Tree(p.right, q.left)     
            return False
        return Tree(root.left, root.right

 

方法二：迭代：

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        if not root: return True
        def Tree(p, q):
            stack = [(q, p)]
            while stack:
                a, b = stack.pop()
                if not a and not b:
                    continue
                if a and b and a.val == b.val:
                    stack.append((a.left, b.right))
                    stack.append((a.right,b.left))
                else:
                    return False
            return True
        return Tree(root.left, root.right)

 方法三：层次遍历

# 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
        ans = [root.left,root.right]
        while ans:
            tmp,n= [],len(ans)
            for i in range(n):
                r = ans.pop(0)
                if r:
                    ans.append(r.left)
                    ans.append(r.right)
                    tmp.append(r.val)
                else:
                    tmp.append(None)
            if tmp != tmp[::-1]:
                return False   
        return True