Given n, how many structurally unique BSTs (binary search trees) that store values 1...n?

Example

Given n = 3, there are a total of 5 unique BST's.

1           3    3       2      1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
分析:

当数组为 1,2,3,4,.. i,.. n时,基于以下原则的BST建树具有唯一性:以i为根节点的树,其左子树由[0, i-1]构成, 其右子树由[i+1, n]构成。那么,左子树由[0 , i -1]组成时,又有多少组合方式呢?这就可以递归了嘛。

 public class Solution {
public int numTrees(int n) {
int[] result = new int[n + ];
result[] = result[] = ; for (int i = ; i <= n; ++i) {
for (int j = ; j < i; ++j) { // j refers to the number of nodes in the left subtree
result[i] += result[j] * result[i - j - ];
}
}
return result[n];
}
}

Unique Binary Search Trees II

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

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
分析:
这题是要求把每一个树都找出来。从上一题我们知道,当root选取i时,它的左子树是节点从1到i -1的数组成的,右子树是i + 1到n的数组成的。那么我们创建一个函数,返回从 start 到end这两个值之间所有树的root.
 /**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @paramn n: An integer
* @return: A list of root
*/
public ArrayList<TreeNode> generateTrees(int n) {
return generate(, n);
} private ArrayList<TreeNode> generate(int start, int end){
ArrayList<TreeNode> treeList = new ArrayList<>(); if(start > end){
treeList.add(null);
return treeList;
} for(int i = start; i <= end; i++){
ArrayList<TreeNode> left = generate(start, i-);
ArrayList<TreeNode> right = generate(i+, end);
for(TreeNode l: left){
for(TreeNode r: right){
TreeNode root = new TreeNode(i);
root.left = l;
root.right = r;
treeList.add(root);
}
}
}
return treeList;
}
}

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