**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 :

` ````
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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