文巾解题 100. 相同的树

Posted 2021-10-12 UQI-LIUWJ

1 题目描述

 

 2 解题思路

2.1 DFS

class Solution:
    def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
        if(p==None and q==None):
            return True
        elif(p==None or q==None):
            return False

        def dfs(p_,q_):
            if(p_==None and q_==None):
                return True
            elif(p_==None or q_==None):
                return False
            elif(p_.val !=q_.val):
                return False
#判断此时的q和p 根节点是否相等
            else:
                return dfs(p_.left,q_.left) and dfs(p_.right,q_.right)
#递归判断p和q的左和右子树       
        return(dfs(p,q))

 2.2 BFS

class Solution:
    def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
        if(p==None and q==None):
            return True
        elif(p==None or q==None):
            return False

        p_=[p]
        q_=[q]
        while(p_ and q_):
            tmp_p=p_.pop(0)
            tmp_q=q_.pop(0)
            if(tmp_p==None and tmp_q==None):
                continue
            elif(tmp_p==None or tmp_q==None):
                return False
            elif(tmp_p.val!=tmp_q.val):
                return False
#判断当前p和q的根节点是否相同
            else:
                p_.append(tmp_p.left)
                q_.append(tmp_q.left)
                p_.append(tmp_p.right)
                q_.append(tmp_q.right)
#将p和q的左右子节点入栈
        return True