95. Unique Binary Search Trees II

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

Constraints:

• 0 <= n <= 8

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

we have nodes from 1 to n, and everyone can be the root, so we from 1 to n pick one as the root and try to find its left childs and right childs. How?

we find its lefttree by build(start, index - 1), right tree by build(index + 1, end), then we know we have leftchild.size() * right.size() combinations, so we make a double for-loop and build every combination and add to the result.

标签：Binary,Search,right,indx,Trees,start,build,null,root
来源： https://www.cnblogs.com/wentiliangkaihua/p/13375469.html