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
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
if p == q:
return True
if p is None or q is None or p.val != q.val:
return False
return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == q) return true;
if (p == null || q == null || p.val != q.val) return false;
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool isSameTree(TreeNode* p, TreeNode* q) {
if (p == q) return true;
if (!p || !q || p->val != q->val) return false;
return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
}
};
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func isSameTree(p *TreeNode, q *TreeNode) bool {
if p == q {
return true
}
if p == nil || q == nil || p.Val != q.Val {
return false
}
return isSameTree(p.Left, q.Left) && isSameTree(p.Right, q.Right)
}