LeetCode 100.Same Tree

LeetCode 100.Same Tree

Description:

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

---

分析：

一道简单的题目，如果两棵树都为NULL，则返回true；否则从两棵树根部开始判断，若相等，则继续判断左子树，再判断右子树，利用递归即可解决。
这道题我是在做测试样例的时候懵了，因为好久没用过c语言指向结构体的指针，一直写不好main函数，所以又重新复习了一下指向结构体的指针，代码不常写真是不行。

代码如下：

#include <iostream>
using namespace std;
/**
 * 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) {
		// return (p != NULL && q != NULL && p->val == q->val 
		// 	&& isSameTree(p->left, q->left) && isSameTree(p->right, q->right)) 
		// 	|| (p == NULL && q == NULL);

		if (p == NULL && q == NULL) return true;
		else if (p != NULL && q != NULL && p-> val == q->val && isSameTree(p->left, q->left) && isSameTree(p->right, q->right)) 
			return true;
		else return false;
	}
};
int main() {
	Solution s;

	int val1;
	cin >> val1;
	TreeNode tree1(val1);

	int left1;
	cin >> left1;
	TreeNode l1(left1);

	int right1;
	cin >> right1;
	TreeNode r1(right1);
	// 构造指向结构体的指针
	TreeNode* p;
	p = &tree1;
	p->left = &l1;
	p->right = &r1;

	int val2;
	cin >> val2;
	TreeNode tree2(val2);
	

	int left2;
	cin >> left2;
	TreeNode l2(left2);

	int right2;
	cin >> right2;
	TreeNode r2(right2);
	// 构造指向结构体的指针
	TreeNode* q;
	q = &tree2;
	q->left = &l2;
	q->right = &r2;
	cout << s.isSameTree(p, q) << endl;
	return 0;
}