95. Unique Binary Search Trees II

与96. Unique Binary Search Trees很相似,

再新建一个function，传入起始位置start和结束位置end，思路虽然知道，但是总是写不出来，应该是敲代码敲的还不多，还不熟悉一些套路

Given an integer 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

import java.util.ArrayList;
import java.util.List;

public class Solution {

public List<TreeNode> generateTrees(int n) {
if(n == 0) return new ArrayList<TreeNode>();

// 套路：不需要其他代码，只要一个helper function
return generateTrees(1, n);
}

private List<TreeNode> generateTrees(int start, int end) {
List<TreeNode> rst = new ArrayList<TreeNode>();
if(start > end) {
rst.add(null);
return rst;
}

if(start == end) {
rst.add(new TreeNode(start));
return rst;
}

for(int i=start; i<=end; i++) {
List<TreeNode> lefts = generateTrees(start, i-1);
List<TreeNode> rights = generateTrees(i+1, end);

// lefts, rights也会有null
for(TreeNode left : lefts) {
for(TreeNode right : rights){
TreeNode head = new TreeNode(i);
head.left = left;
head.right = right;
rst.add(head);
}
}
}

return rst;
}

}