22 Generate Parentheses

定义问题：左括号，右括号需要匹配，也就是必须相等

[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]

n = 3 ,说明需要3个左括号，3个右括号

left : 没有匹配的“（”
right:
dfs(int left, int right, char* str, char*result, int returnSize, int n)

left-1： 添加"("
right-1: 添加")"

二叉递归：[30,12,20,03,11,11,#,#,02,02,10,02,10]

recursion()
{
     
    if (end_condition)
{

              solve;     
     
    }
else
{     
        //在将问题转换为子问题描述的每一步，都解决该步中剩余部分的问题。
          
        if()
            recursion();
        if()
            recursion();
    }

}

#define SIZE 10000

void dfs(int left, int right, char* str, char**result, int* returnSize, int n) {
    if((left == 0) && (right == 0))
        result[(*returnSize)++] = str;
    else {
        char* newStr = (char *)malloc(sizeof(char) * (2*n+1));
        if(left > 0) {
            strcpy(newStr, str);
            dfs(left-1, right+1, strcat(newStr, "("), result, returnSize, n);
        }
        if(right > 0) {
            strcpy(newStr, str);
            dfs(left, right-1, strcat(newStr, ")"), result, returnSize, n);
        }
    }
}

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
char** generateParenthesis(int n, int* returnSize) {
    char** result = (char **)malloc(sizeof(char *) * SIZE);
    dfs(n, 0, "", result, returnSize, n);
    return result;
}