NO.1

思路：递归。

代码实现：

class Solution {
public:bool IsBalanced_Solution(TreeNode* pRoot) {return dfs(pRoot)!=-1;}int dfs(TreeNode* root){if(root==nullptr) return 0;int left=dfs(root->left);if(left==-1) return -1;int right=dfs(root->right);if(right==-1) return -1;return abs(left-right)<=1?max(left,right)+1:-1;}
};

NO.2

算法思路：
⼆维前缀和矩阵的应⽤。
a. 初始化⼆维前缀和矩阵；
b. 枚举所有的⼦矩阵，求出最⼤⼦矩阵。

代码实现：

#include <iostream>
using namespace std;
const int N = 110;
int n;
int dp[N][N];
int main()
{int x;cin >> n;for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){cin >> x;dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + x;}}int ret = -127 * N;for (int x1 = 1; x1 <= n; x1++){for (int y1 = 1; y1 <= n; y1++){for (int x2 = x1; x2 <= n; x2++){for (int y2 = y1; y2 <= n; y2++){ret = max(ret, dp[x2][y2] - dp[x1 - 1][y2] - dp[x2][y1 -1] + dp[x1 - 1][y1 - 1]);}}}}cout << ret << endl;return 0;
}

NO.3

思路：滑动窗口，分别统计字符串中和窗口内0和1的数量，当窗口内字符串的数量为原字符串的一半，且该窗口内0和1的数量是外面字符串中0和1数量的一半时，就有两种方法。

代码实现：

#include <iostream>
#include <string>using namespace std;
int n;
string s;
int main()
{cin >> n >> s;int sum[2] = { 0 }; // 统计字符串中所有 0 和 1 的个数for (auto ch : s){sum[ch - '0']++;}int left = 0, right = 0, ret = 0, half = n / 2;int count[2] = { 0 }; // 统计窗⼝内 0 和 1 的个数while (right < n - 1) // 细节问题{count[s[right] - '0']++;while (right - left + 1 > half){count[s[left++] - '0']--;}if (right - left + 1 == half){if (count[0] * 2 == sum[0] && count[1] * 2 == sum[1]){ret += 2;}}right++;}cout << ret << endl;return 0;
}