leetcode -- 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

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as

"{1,2,3,#,#,4,#,#,5}"

.
[解题思路]

划分成左右子树分别进行构造，当左右子树所有可能情况都构造完毕后，加上node即可

这里根节点可能情况为1,2,....,n

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; left = null; right = null; }
* }
*/
public class Solution {
public ArrayList<TreeNode> generateTrees(int n) {
// Start typing your Java solution below
// DO NOT write main() function
if(n == 0){
return generate(1, 0);
}
return generate(1, n);
}

public ArrayList<TreeNode> generate(int start, int end){
ArrayList<TreeNode> subTree = new ArrayList<TreeNode>();
if(start > end){
subTree.add(null);
return subTree;
}

for(int i = start; i <= end; i++){
ArrayList<TreeNode> leftSubTree = generate(start, i - 1);
ArrayList<TreeNode> rightSubTree = generate(i + 1, end);
for(int j = 0; j < leftSubTree.size(); j++){
for(int k = 0; k < rightSubTree.size(); k++){
TreeNode node = new TreeNode(i);
node.left = leftSubTree.get(j);
node.right = rightSubTree.get(k);
subTree.add(node);
}
}
}
return subTree;
}
}