Unique Binary Search Trees II LeetCode Solution

Problem – Unique Binary Search Trees II LeetCode Solution

Given an integer n, return all the structurally unique BST’s (binary search trees), which has exactly n nodes of unique values from 1 to n. Return the answer in any order.

Example 1:

Input: n = 3
Output: [[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]

Example 2:

Input: n = 1
Output: [[1]]

Constraints:

  • 1 <= n <= 8

Unique Binary Search Trees II LeetCode Solution in Java

public static List<TreeNode> generateTrees(int n) {
    List<TreeNode>[] result = new List[n + 1];
    result[0] = new ArrayList<TreeNode>();
    if (n == 0) {
        return result[0];
    }

    result[0].add(null);
    for (int len = 1; len <= n; len++) {
        result[len] = new ArrayList<TreeNode>();
        for (int j = 0; j < len; j++) {
            for (TreeNode nodeL : result[j]) {
                for (TreeNode nodeR : result[len - j - 1]) {
                    TreeNode node = new TreeNode(j + 1);
                    node.left = nodeL;
                    node.right = clone(nodeR, j + 1);
                    result[len].add(node);
                }
            }
        }
    }
    return result[n];
}

private static TreeNode clone(TreeNode n, int offset) {
    if (n == null) {
        return null;
    }
    TreeNode node = new TreeNode(n.val + offset);
    node.left = clone(n.left, offset);
    node.right = clone(n.right, offset);
    return node;
}

Unique Binary Search Trees II LeetCode Solution in Python

def generateTrees(self, n):
    def node(val, left, right):
        node = TreeNode(val)
        node.left = left
        node.right = right
        return node
    def trees(first, last):
        return [node(root, left, right)
                for root in range(first, last+1)
                for left in trees(first, root-1)
                for right in trees(root+1, last)] or [None]
    return trees(1, n)

Unique Binary Search Trees II LeetCode Solution in C++

class Solution {
    public:
        TreeNode* clone(TreeNode* root){
            if(root == nullptr)
                return nullptr;
            TreeNode* newroot = new TreeNode(root->val);
            newroot->left = clone(root->left);
            newroot->right = clone(root->right);
            return newroot;
        }
        vector<TreeNode *> generateTrees(int n) {
            vector<TreeNode *> res(1,nullptr);
            for(int i = 1; i <= n; i++){
                vector<TreeNode *> tmp;
                for(int j = 0; j<res.size();j++){
                    TreeNode* oldroot = res[j];
                    TreeNode* root = new TreeNode(i);
                    TreeNode* target = clone(oldroot);
                    root->left = target;
                    tmp.push_back(root);
                   
                    if(oldroot!=nullptr){
                        TreeNode* tmpold = oldroot;
                        while(tmpold->right!=nullptr){
                            TreeNode* nonroot = new TreeNode(i);
                            TreeNode *tright = tmpold->right;
                            tmpold->right = nonroot;
                            nonroot->left = tright;
                            TreeNode *target = clone(oldroot);
                            tmp.push_back(target);
                            tmpold->right = tright;
                            tmpold = tmpold->right;
                        }
                        tmpold->right = new TreeNode(i);
                        TreeNode *target = clone(oldroot);
                        tmp.push_back(target);
                        tmpold->right = nullptr;
                    }
                }
                res=tmp;
            }
            return res;
        }
    };
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