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
这个题目利用深度搜索去做,因为只要发现有一个分支不对称,那么就可以判断这两个二叉树不相同
#include <iostream>
using namespace std;
//////链表的节点创建,创建一个节点/////////
struct TreeNode {
int val;
TreeNode *lp;
TreeNode *rp;
TreeNode();
TreeNode(int a){val=a;lp=NULL;rp=NULL;}; ////这里创建一个结构体的构造函数,很方便,初始化的时候直接进行初始化
};
bool isSameTree(TreeNode* p, TreeNode* q) {
if (p == NULL && q == NULL) return true; ////当搜索到最底层的时候,返回true
if (p == NULL || q == NULL) return false;
if (p->val != q->val) return false; /////
return isSameTree(p->lp, q->lp) && isSameTree(p->rp, q->rp); ///这种情况实际上是p和q的val相同的情况,那么就继续比较他们的左右子树,
}
int main() {
TreeNode *a = new TreeNode(1);
TreeNode *b = new TreeNode(2);
TreeNode *c = new TreeNode(3);
TreeNode *d = new TreeNode(4);
TreeNode *e = new TreeNode(5);
TreeNode *f = new TreeNode(6);
TreeNode *a1 = new TreeNode(1);
TreeNode *b1 = new TreeNode(2);
TreeNode *c1= new TreeNode(3);
TreeNode *d1 = new TreeNode(4);
TreeNode *e1 = new TreeNode(5);
TreeNode *f1 = new TreeNode(6);
a->lp = b;
a->rp = c;
b->lp = d;
b->rp = e;
c->lp = f;
a1->lp = b1;
a1->rp = c1;
b1->lp = d1;
// b1->rp = e1;
c1->lp = f1;
isSameTree(a,a1);
return 0;
}
这里我做的二叉树是这样的:
输入两个二叉树:
1 1 / \ / \ 2 3 2 3
/ \ / / /
4 5 6 4 6
通过debug发现,一次搜索的顺序是1,2,4,5 然后就结束了,
说明当 return A&&B的时候,只要确认A是false,那么就没必要再计算B了,但是如果A是true的话,此时无法确认返回值,则需要继续计算B,正是因为语言有这样的功能,才正真实现了,深度搜索。