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【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 }