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(, 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 - );
             List<TreeNode> rTree = dfs(i + , 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