Unique Binary Search Trees II

Unique Binary Search Trees II

Description:

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

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

分析：

思路跟Unique Binary Search Trees很像。

考虑根节点时i的时候，则比i小的都在左子树，比i大的都在右子树。但是此时左、右子树都是多种情况的，所以还需要对左、右子树的所有情况进行组合。

注意：这个博客指出用指针的话可以节省空间，没有仔细去看。

参考：

Code：

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/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @paramn n: An integer
     * @return: A list of root
     */
    vector<TreeNode *> generateTrees(int n) {
        // write your code here
        return dfs(1, n);
    }
    
    vector<TreeNode *>  dfs(int min, int max) {
        vector<TreeNode *> rs;
        if (min > max) {
            rs.push_back(nullptr);
            return rs;
        }
        
        for (int i = min; i <= max; i++) {
            vector<TreeNode *> left = dfs(min, i-1);
            vector<TreeNode *> right = dfs(i+1, max);
            
            for(int j = 0; j < left.size(); j++) {
                for (int k = 0; k < right.size(); k++) {
                    TreeNode *root = new TreeNode(i);
                    root->left = left[j];
                    root->right = right[k];
                    rs.push_back(root);
                }
            }
        }
        
        return rs;
    }
};