LeetCode100——Same Tree

本文探讨了如何检查两棵二叉树是否完全相同,包括结构和节点值。通过递归和非递归方法实现,提供了Python代码示例,并对比了两种方法的运行效率。

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题目链接

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example 1:

Input:     1         1
          / \       / \
         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true

Example 2:

Input:     1         1
          /           \
         2             2

        [1,2],     [1,null,2]

Output: false

Example 3:

Input:     1         1
          / \       / \
         2   1     1   2

        [1,2,1],   [1,1,2]

Output: false

看到题目的第一时间是没有思路的,可能用深搜、广搜?后来想了一下,不就是比较每一个节点的值,然后再比较它们的左右子树是否相同吗?因此,想到了递归算法。代码如下

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

        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)

测评的结果

Runtime: 36 ms, faster than 71.97% of Python3 online submissions for Same Tree.
Memory Usage: 13.1 MB, less than 5.74% of Python3 online submissions for Same Tree.

既然有递归算法,那么我们也尝试一下非递归算法

class Solution:
    def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
        stack = [(p, q)]

        while(len(stack)!= 0):
            temp_p, temp_q = stack.pop()
            if temp_p == None and temp_q == None:
                continue
            if temp_p == None and temp_q != None or temp_p != None and temp_q == None:
                return False
            if temp_p.val != temp_q.val:
                return False

            stack.append((temp_p.left, temp_q.left))
            stack.append((temp_p.right, temp_q.right))
        return True

测评结果

Runtime: 40 ms, faster than 39.18% of Python3 online submissions for Same Tree.
Memory Usage: 13.2 MB, less than 5.74% of Python3 online submissions for Same Tree.

 

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