题目描述
请实现一个函数,用来判断一颗二叉树是不是对称的。注意,如果一个二叉树同此二叉树的镜像是同样的,定义其为对称的。
思路(参考
1.只要pRoot.left和pRoot.right是否对称即可
2.左右节点的值相等且对称子树(left.left, right.right );(left.rigth,right.left)也对称
代码
1.递归
#递归,两个树对称的条件是左子树的左子树和右子树的右子树相同;左子树的右子树和右子树的左子树相同 class Solution: def isSymmetrical(self, pRoot): # write code here if pRoot==None: return True left = pRoot.left right = pRoot.right return self.isSymmetricalHelper(left,right) def isSymmetricalHelper(self,left,right): if left==None and right == None: return True if left==None or right == None: return False if left.val==right.val and self.isSymmetricalHelper(left.left,right.right) and self.isSymmetricalHelper(left.right,right.left): return True return False
2.非递归
使用栈
# -*- coding:utf-8 -*- # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def isSymmetrical(self, pRoot): # write code here #使用栈 if pRoot == None: return True stack = [(pRoot.left,pRoot.right)] while(stack): l,r = stack.pop() if l==None and r==None: continue if l==None or r==None: return False if l.val!=r.val: return False stack.append((l.left,r.right)) stack.append((l.right,r.left)) return True