树的练习（4）

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        def recur(L, R):
            if not L and not R: return True
            if not L or not R or L.val != R.val: return False
            return recur(L.left, R.right) and recur(L.right, R.left)
        return recur(root.left, root.right) if root else True
如果上面的不容易理解，可以看下面的

# 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
        def recur(L, R):
            if not L and not R: return True
            if not L or not R or L.val != R.val: return False
            return recur(L.left, R.right) and recur(L.right, R.left)
        return recur(root.left, root.right) 

迭代
class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        if root is None:
            return True
        q = []
        q.append(root.left)
        q.append(root.right)
        while len(q)!=0:
            A = q.pop(0)
            B = q.pop(0)
            if A == None and B == None:
                continue
            if A == None or B == None:
                return False
            if A.val != B.val:
                return False
            q.append(A.left)
            q.append(B.right)
            q.append(A.right)
            q.append(B.left)
        return True 

将上一题。改进：

class Solution:
    def isSymmetric_queue(self, root: TreeNode) -> bool:
        """
        辅助队列法
        """
        if not root:
            return True
        queue = [(root.left,root.right)]
        while queue:
            l, r = queue.pop(0)
            if not l and not r: # 如果左右子节点为空，跳过后面步骤
                continue
            if not l or not r: # 如果左右子节点任意一个为空，则表示不对称
                return False
            if l.val != r.val: # 如果左右子节点的值不等，则表示不对称
                return False
            queue.append((l.left, r.right)) # 将左子树的左节点，和右子树的右节点放入队列
            queue.append((l.right, r.left))
        return True