Leetcode-100. 相同的树

版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/changjiale110/article/details/85374440

给定两个二叉树,编写一个函数来检验它们是否相同。

如果两个树在结构上相同,并且节点具有相同的值,则认为它们是相同的。

示例 1:

输入:       1         1
          / \       / \
         2   3     2   3

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

输出: true
示例 2:

输入:      1          1
          /           \
         2             2

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

输出: false
示例 3:

输入:       1         1
          / \       / \
         2   1     1   2

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

输出: false

c++写法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        if (p==NULL&&q==NULL)
            return true;
        else if (p==NULL||q==NULL)
            return false;
        else if (p->val==q->val)
        {
            bool left = isSameTree(p->left,q->left);
            bool right = isSameTree(p->right,q->right);
            return (left&&right);
        }
        else 
        {
            return false;
        }
        
        
    }
};

py写法

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

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

猜你喜欢

转载自blog.csdn.net/changjiale110/article/details/85374440