95. Unique Binary Search Trees II (M)

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

题意

给定n个整数1，2，…，n，用这些整数构成二叉搜索树BST的结点。返回所有可以形成的BST。

思路

96. Unique Binary Search Trees 威力加强版。同样的思想，每次固定一个根结点，生成所有可能的BST。利用分治法和递归进行处理。

代码实现

class Solution {
    public List<TreeNode> generateTrees(int n) {
        if (n == 0) {
            return new ArrayList<>();
        }
        return solve(1, n);
    }
    private List<TreeNode> solve(int start, int end) {
        List<TreeNode> list = new ArrayList<>();
        if (start > end) {
            list.add(null);
        }
        for (int i = start; i <= end; i++) {
            List<TreeNode> left = solve(start, i - 1);
            List<TreeNode> right = solve(i + 1, end);
            for (TreeNode p : left) {
                for (TreeNode q : right) {
                    TreeNode root = new TreeNode(i);
                    root.left = p;
                    root.right = q;
                    list.add(root);
                }
            }
        }
        return list;
    }
}