Unique Binary Search Trees Series

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Unique Binary Search Trees Series

Question

Given n, how many structurally unique BST’s (binary search trees) that store values 1…n?

For example,

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

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1         3     3      2      1
 \       /     /      / \      \
  3     2     1      1   3      2
 /     /       \                 \
2     1         2                 3

Analysis

LeetCode Discuss Detailed explanation

对于数字范围1-n，可能任意选择其中的x作为树的根，以x为根的树的种类数是[1,x-1]和[x+1,n]可形成的左右子树个数的乘积和。

Code

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public class Solution {
    public int numTrees(int n) {
        int[] numTrees=new int[n+1];
        numTrees[0]=1;
        numTrees[1]=1;
        for(int i=2;i<=n;i++){
            for(int j=1;j<=i;j++){
                numTrees[i]+=numTrees[j-1]*numTrees[i-j];
            }
        }
        return numTrees[n];
    }
}

Unique Binary Search Trees Series II

Question

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.

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5
1         3     3      2      1
 \       /     /      / \      \
  3     2     1      1   3      2
 /     /       \                 \
2     1         2                 3

Analysis

LeetCode Discuss Explanation

原理同上，只不过得到的不是树的个数而是左右子树的节点，然后利用循环进行组合

Code

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<TreeNode> generateTrees(int n) {
        List<TreeNode> res=new ArrayList<TreeNode>();
        if(n==0)    return res;
        return helper(1,n);
    }
   
    private List<TreeNode> helper(int start, int end){
        List<TreeNode> res=new ArrayList<TreeNode>();
        if(start>end){
            res.add(null);
            return res;
        }
        
        for(int i=start;i<=end;i++){
            List<TreeNode> left=helper(start,i-1);
            List<TreeNode> right=helper(i+1,end);
            
            for(TreeNode lchild:left){
                for(TreeNode rchild:right){
                    TreeNode root=new TreeNode(i);
                    root.left=lchild;
                    root.right=rchild;
                    res.add(root);
                }
            }
        }
        return res;
    }
}