[LeetCode] Unique Binary Search Trees II

最新推荐文章于 2020-12-09 16:51:01 发布

最新推荐文章于 2020-12-09 16:51:01 发布 · 80 阅读

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本文介绍了一种算法，用于生成所有可能的独特二叉搜索树（BSTs），这些树能存储从1到n的值。通过递归地构造左右子树来确保生成的BST结构各异。示例展示了当n为3时，可以生成五种不同的BST结构。

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * 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 *> generateTrees(int n) {
        
        return getTrees(1, n);
    }
    vector<TreeNode*> getTrees(int start, int end) {
        vector<TreeNode* > res;
        if ( start > end ) {
            res.push_back(NULL);
            return res;
        }
        for (int i = start; i <= end; i++) {
            vector<TreeNode*> left = getTrees(start, i-1);
            vector<TreeNode*> right = getTrees(i+1, end);
            
            for (int l = 0; l < left.size(); l++) {
                for (int r = 0; r < right.size() ;r++) {
                    TreeNode* cur = new TreeNode(i);
                    cur->left = left[l];
                    cur->right = right[r];
                    res.push_back(cur);
                }
            }
        }
    }
};