[leetcode:python]100.Same Tree

最新推荐文章于 2022-10-14 10:51:19 发布

AliceWanderAI

最新推荐文章于 2022-10-14 10:51:19 发布

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分类专栏：算法文章标签： python leetcode

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本文介绍了三种不同的方法来判断两棵二叉树是否等价，并对比了它们的性能差异。方法一是通过递归的方式检查节点值是否相同，方法二在方法一的基础上调整了返回False的时机，从而提高了性能；方法三进一步优化了递归条件，达到了最佳性能。

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题目：
Given two binary trees, write a function to check if they are equal or not.

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

方法一：性能59ms

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSameTree(self, p, q):
        """
        :type p: TreeNode
        :type q: TreeNode
        :rtype: bool
        """
        if p == None and q == None:
            return True
        if p and q and p.val == q.val:
            return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)

        else:
            return False

方法二：性能38ms

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSameTree(self, p, q):
        """
        :type p: TreeNode
        :type q: TreeNode
        :rtype: bool
        """
        if p == None and q == None:
            return True
        if p and q and p.val == q.val:
            return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
        return False

这两个方法的唯一区别就是最后那个return False的位置，造成的性能差异还是有点大的。。。值得注意。

方法三：性能32ms

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSameTree(self, p, q):
        """
        :type p: TreeNode
        :type q: TreeNode
        :rtype: bool
        """
        if p and q:
           return (p.val==q.val) and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
        elif p==q:
            return True
        else:
            return False