LeetCode——Unique Binary Search Trees II

Question

Given an integer 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

Solution

分治的思想，分别求出1～n个节点为根节点构成的BST.

例如 k, 1<= k <= n; 为根节点的问题可以划分为，左子树和右子树，左子树中的节点值为1_{k-1，然后1}k-1分别为根节点，同理右子树。右子树中的节点值为k+1_{n，然后k+1}n分别为根节点，以此类推。

Code

/**
 * Definition for a binary tree node.
 * 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) {
        if (n <= 0)
            return vector<TreeNode*>();
        else
            return generateSubTree(1, n);
    }
    vector<TreeNode*> generateSubTree(int s, int e) {
        vector<TreeNode*> res;
        if (s > e) {
            res.push_back(NULL);
            return res;
        }
        for (int i = s; i <= e; i++) {
            vector<TreeNode*> left = generateSubTree(s, i - 1);
            vector<TreeNode*> right = generateSubTree(i + 1, e);
            
            // 所有组合 ＝ size(left) * size(right)
            for (TreeNode* p : left) {
                for (TreeNode* q : right) {
                    struct TreeNode* root = new TreeNode(i);
                    root->left = p;
                    root->right = q;
                    res.push_back(root);
                }
            }
        }
        return res;
    }
};