【LeetCode 95】 Unique Binary Search Trees II

題目描述

Given an integer n, generate all structurally unique BST’s (binary search trees) that store values 1 … n.

Example:

Input: 3
Output:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]

Explanation:

The above output corresponds to the 5 unique BST's shown below:

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

Constraints:
0 <= n <= 8

思路

i爲頂點的時候，1~i-1在左子樹，i+1 ~ n在右子樹。遞歸建立所有子樹。

代碼

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        if (n <= 0) return {};
        return generateSubTree(1, n);
    }
    
    vector<TreeNode*> generateSubTree(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 = generateSubTree(start, i-1);
            vector<TreeNode*> right = generateSubTree(i+1, end);
            for (auto l : left) {
                for (auto r : right) {
                    TreeNode* root = new TreeNode(i);
                    root->left = l;
                    root->right = r;
                    res.push_back(root);
                }
            }
        }
        
        return res;
    }
};