[leetcode]Unique Binary Search Trees II

最新推荐文章于 2025-09-09 19:16:33 发布

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最新推荐文章于 2025-09-09 19:16:33 发布

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CC 4.0 BY-SA版权

文章标签： c/c++

原文链接：http://www.cnblogs.com/dyllove98/p/3184817.html

本文提供了一种方法来生成所有结构独特的BST（二叉搜索树），对于给定的整数n，生成包含数值1到n的所有可能的BST结构。通过递归调用函数来构建树的不同部分，并将它们组合在一起，最终返回所有可能的BST实例。

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.

For example,
Given n = 3, your program should return all 5 unique BST's shown below.

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode *> generateSub(int N, int start){
        if(N == 0){
            //   return vector<TreeNode *>(1, NULL);
            vector<TreeNode*> tmp;
            tmp.push_back(NULL);
            return tmp;
        }
         
        
       vector<TreeNode *> curLeft, curRight, result;
        for(int n = 1; n <= N; n++){
            int rootval = n + start - 1;
            
            curLeft = generateSub(n-1, start);
            curRight = generateSub(N-n, rootval+1);
            
            for(int i = 0; i < curLeft.size(); i++){
                for(int j = 0; j < curRight.size(); j++){
                    TreeNode *root = new TreeNode(rootval);
                    root -> left = curLeft[i];
                    root -> right = curRight[j];
                    
                    result.push_back(root);
                }
            } 
        }
    
        return result;
    }

    vector<TreeNode *> generateTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return generateSub(n, 1);
        
    }
};

转载于:https://www.cnblogs.com/dyllove98/p/3184817.html