LeetCode: Unique Binary Search Trees II 解题报告

Unique Binary Search Trees II

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.

 

Hide Tags Tree Dynamic Programming

SOLUTION 1:

使用递归来做。

1. 先定义递归的参数为左边界、右边界，即1到n.

2. 考虑从left, 到right 这n个数字中选取一个作为根，余下的使用递归来构造左右子树。 

3. 当无解时，应该返回一个null树，这样构造树的时候，我们会比较方便，不会出现左边解为空，或是右边解为空的情况。

4. 如果说左子树有n种组合，右子树有m种组合，那最终的组合数就是n*m. 把这所有的组合组装起来即可

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; left = null; right = null; }
 * }
 */
public class Solution {
    public List<TreeNode> generateTrees(int n) {
        // 0.07
        return dfs(1, n);
    }
    
    public List<TreeNode> dfs(int left, int right) {
        List<TreeNode> ret = new ArrayList<TreeNode>();
        
        // The base case;
        if (left > right) {
            ret.add(null);
            return ret;
        }
        
        for (int i = left; i <= right; i++) {
            List<TreeNode> lTree = dfs(left, i - 1);
            List<TreeNode> rTree = dfs(i + 1, right);
            for (TreeNode nodeL: lTree) {
                for (TreeNode nodeR: rTree) {
                    TreeNode root = new TreeNode(i);
                    root.left = nodeL;
                    root.right = nodeR;
                    ret.add(root);
                }
            }
        }
        
        return ret;
    }
}

CODE:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/72f914daab2fd9e2eb8a63e4643297d78d4fadaa/tree/GenerateTree2.java