Lowest Common Ancestor - Amazon Top Interview Questions


Problem Statement :


Given a binary tree root, and integers a and b, find the value of the lowest node that has a and b as descendants. A node can be a descendant of itself.

All nodes in the tree are guaranteed to be unique.

Constraints

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

Example 1

Input

root = [0, [1, null, null], [2, [6, [3, null, null], [4, null, null]], [5, null, null]]]

a = 3

b = 5

Output

2

Example 2

Input

root = [0, [1, null, null], [2, [6, [3, null, null], [4, null, null]], [5, null, null]]]

a = 6

b = 4

Output

6


Solution :



title-img 




                        Solution in C++ :

int solve(Tree* root, int a, int b) {
    if (root == NULL) return INT_MIN;
    if (root->val == a or root->val == b) return root->val;

    int left = solve(root->left, a, b);
    int right = solve(root->right, a, b);

    if (left != INT_MIN and right != INT_MIN) {
        return root->val;
    }

    return left != INT_MIN ? left : right;
}
                    


                        Solution in Java :

import java.util.*;

/**
 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
 */
class Solution {
    public Tree lca(Tree root, int a, int b) {
        if (root == null) {
            return null;
        }

        if (root.val == a || root.val == b) {
            return root;
        }

        Tree left = lca(root.left, a, b);
        Tree right = lca(root.right, a, b);

        if (left != null && right != null) {
            return root;
        }

        if (left != null) {
            return left;
        }

        if (right != null) {
            return right;
        }

        return null;
    }

    public int solve(Tree root, int a, int b) {
        return lca(root, a, b).val;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, root, a, b):
        nodes = {}
        count = 0

        def dfs(root, st):
            nonlocal nodes, count
            if count < 2:
                if not root:
                    return
                if root.left:
                    dfs(root.left, st + [root.left.val])
                if root.val in [a, b]:
                    count += 1
                nodes[root.val] = st
                if root.right:
                    dfs(root.right, st + [root.right.val])
            else:
                return

        dfs(root, [root.val])
        st1, st2 = nodes[a], nodes[b]
        for i in range(min(len(st1), len(st2))):
            if st1[i] == st2[i]:
                ans = st1[i]
        return ans


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