95. Unique Binary Search Trees II

时间：2018-09-10 18:33:19 阅读：262 评论：0 收藏：0 [点我收藏+]

标签：tac 2.0 img inline 3.4 bubuko array 左右 col

题目

Given an integer n, generate all structurally unique BST‘s (binary search trees) that store values 1 ... n.

Example:

Input: 3
Output:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST‘s shown below:

   1         3     3      2      1
    \       /     /      / \           3     2     1      1   3      2
    /     /       \                    2     1         2                 3

思路

记f_n为1至n组成的BST的集合。对该BST，根的值有n种情况。对每一个根值i，左树为1至i-1，即问题f_i-1。右树为i+1至n，即问题f_n-i，但每个节点要+i。即

f_n = ∑ f_i-1 * f_n-i (i=1...n)

注：此处答案要求是一个List，所以*不代表数学的乘积而是将左右树做交叉结合

代码

 1 class Solution {
 2     public List<TreeNode> generateTrees(int n) {
 3         List<TreeNode>[] dp = new List[n + 1];
 4         dp[0] = new ArrayList<>();
 5         if(n == 0) return dp[0];
 6         dp[0].add(null);
 7         
 8         for(int i = 1; i <= n; i++){
 9             dp[i] = new ArrayList<>();
10             for(int j = 1; j <= i; j++){
11                 int left = j - 1, right = i - j;
12                 for(TreeNode leftTree : dp[left])
13                     for(TreeNode rightTree : dp[right]){
14                         TreeNode root = new TreeNode(j);
15                         root.left = clone(leftTree, 0);
16                         root.right = clone(rightTree, j);
17                         dp[i].add(root);
18                     }
19             }
20         }
21         return dp[n];
22     }
23     
24     private TreeNode clone(TreeNode x, int offset){
25         if(x == null)
26             return null;
27         TreeNode result = new TreeNode(x.val + offset);
28         result.left = clone(x.left, offset);
29         result.right = clone(x.right, offset);
30         return result;
31     }
32 }