Lowest Common Ancestor of List of Values - Amazon Top Interview Questions


Problem Statement :


Given a binary tree root containing unique values and a list of unique integers values, return the lowest common ancestor node that contains all values in values. You can assume that a solution exists.

Constraints

n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [1, [2, null, null], [3, [4, [6, null, null], [7, null, null]], [5, null, null]]]

values = [5, 6, 7]

Output

[3, [4, [6, null, null], [7, null, null]], [5, null, null]]


Solution :



title-img 




                        Solution in C++ :

Tree* tra(Tree* root, unordered_set<int>& s) {
    if (!root) return root;
    if (s.find(root->val) != s.end()) return root;
    Tree* x = tra(root->left, s);
    Tree* y = tra(root->right, s);
    if (x and y) return root;
    return x ? x : y;
}
Tree* solve(Tree* root, vector<int>& values) {
    unordered_set<int> s(values.begin(), values.end());
    return tra(root, s);
}
                    


                        Solution in Java :

import java.util.*;

/**
 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
 */
class Solution {
    private int[] values;
    private Set<Integer> set = new HashSet();
    public Tree solve(Tree root, int[] values) {
        this.values = values;
        Queue<Tree> q = new LinkedList();
        Tree ret = null;
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            for (int x = 0; x < size; x++) {
                Tree temp = q.poll();
                fillSet();
                recurse(temp);
                if (!set.isEmpty())
                    continue;
                ret = temp;
                if (temp.left != null) {
                    q.offer(temp.left);
                }
                if (temp.right != null) {
                    q.offer(temp.right);
                }
            }
        }
        return ret;
    }
    public void recurse(Tree root) {
        if (root == null || set.isEmpty())
            return;
        if (set.contains(root.val)) {
            set.remove(root.val);
        }
        recurse(root.left);
        recurse(root.right);
    }
    public void fillSet() {
        for (int v : values) {
            set.add(v);
        }
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, root, values):
        S = set(values)

        def dfs(node):
            if None == node:
                return None

            if node.val in S:
                return node

            left = dfs(node.left)
            right = dfs(node.right)

            if left and right:
                return node
            else:
                return left or right

        return dfs(root)


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