LeetCode之Generate Parentheses（C++）

来源：互联网发布：淘宝上哪个符咒是真的编辑：程序博客网时间：2024/05/01 12:58

题目：Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"

目标：生成正确的括号对

数据结构：采用二叉树结构，每个节点的value是“（”或“）”，节点结构包括左右孩子节点，父节点，val，栈顶节点，剩下可使用的“（”的数量count，代码表示如下：

        struct TreeNode{TreeNode* left,*right,*parent,*top;char val;int count;TreeNode(char v):val(v),top(NULL),left(NULL),right(NULL),parent(NULL),count(0){};TreeNode(char v,TreeNode* p):val(v),top(NULL),left(NULL),right(NULL),parent(p),count(0){};};

算法思路：当当前节点的栈顶节点为“（”时，可以插入右子节点“）”，此时若count大于0，还可以插入左子节点“（”，当栈顶节点为空时，且count大于0，那么可以插入左子节点“（”，否则只能返回。以上插入完子节点后均可以继续递归。完整代码如下：

class Solution {public:struct TreeNode{TreeNode* left,*right,*parent,*top;char val;int count;TreeNode(char v):val(v),top(NULL),left(NULL),right(NULL),parent(NULL),count(0){};TreeNode(char v,TreeNode* p):val(v),top(NULL),left(NULL),right(NULL),parent(p),count(0){};};void generateTree(TreeNode* node){if(node->top){if(node->count>0){node->left=new TreeNode('(',node);node->left->count=node->count-1;node->left->top=node->left;generateTree(node->left);node->right=new TreeNode(')',node);node->right->count=node->count;if(node->top->parent)node->right->top=node->top->parent->top;generateTree(node->right);}else{node->right=new TreeNode(')',node);node->right->count=node->count;if(node->top->parent)node->right->top=node->top->parent->top;generateTree(node->right);}}else{if(node->count>0){node->left=new TreeNode('(',node);node->left->count=node->count-1;node->left->top=node->left;generateTree(node->left);}else{return;}}}void traverseTree(TreeNode* node,vector<string>& results,string upStr){if(!node->left&&!node->right){ostringstream ostr;ostr<<upStr;ostr.put(node->val);results.push_back(ostr.str());return;}if(node->left){ostringstream ostr;ostr<<upStr;ostr.put(node->val);traverseTree(node->left,results,ostr.str());}if(node->right){ostringstream ostr;ostr<<upStr;ostr.put(node->val);traverseTree(node->right,results,ostr.str());}}    vector<string> generateParenthesis(int n) {        vector<string> results;if(n==0)return results;TreeNode* head=new TreeNode('(');head->count=n-1;head->top=head;generateTree(head);traverseTree(head,results,"");return results;    }};

笔者认为数据结构设计的有些冗余，而且两次使用递归（生成括号和遍历括号树），希望各位技术大牛加以指正，万分感谢。

0 0