**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 :

` ````
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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