【leetcode刷题笔记】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

---

题解：递归的枚举1~n的每个节点为根节点，然后递归的利用它左边的节点构造左子树，放在一个list里面；再利用它右边的节点构造右子树，也放在一个list里面；最终枚举两个list里面的左子树和右子树，构建一棵树。

代码如下：

 1 /**

 2  * Definition for binary tree

 3  * public class TreeNode {

 4  *     int val;

 5  *     TreeNode left;

 6  *     TreeNode right;

 7  *     TreeNode(int x) { val = x; left = null; right = null; }

 8  * }

 9  */

10 public class Solution {

11     private ArrayList<TreeNode> generate(int start,int end){

12         ArrayList<TreeNode> answer = new ArrayList<TreeNode>();

13         if(start > end){

14             answer.add(null);

15             return answer;

16         }

17         

18         //for every node from start to right,make it as tree root and recursively build its left and right child tree

19         for(int i = start; i <= end;i++){

20             ArrayList<TreeNode> left = generate(start, i-1);

21             ArrayList<TreeNode> right = generate(i+1, end);

22             for(TreeNode l:left){

23                 for(TreeNode r:right){

24                     TreeNode root = new TreeNode(i);

25                     root.left = l;

26                     root.right = r;

27                     answer.add(root);

28                 }

29             }

30             

31         }

32         

33         return answer;

34         

35     }

36     public List<TreeNode> generateTrees(int n) {

37         return generate(1,n);

38     }

39 }