Learn and practice the Tree Depth-First Search and Tree Breadth-First Search patterns to solve coding interview problems.

Speedrun the LinkedIn Coding Interview: Top Problems in Java

Trees play a pivotal role in organizing hierarchical data, optimizing searches, and solving complex problems. In this module, we’ll learn to choose the appropriate tree for a given problem and apply the more appropriate of the two major traversal techniques, breadth-first or depth-first, to solve it.

Trees

# Statement

Given the root node of a binary tree with $n$ nodes, your task is to find the lowest common ancestor of two of its nodes, `p` and `q`.

> **Note:** The lowest common ancestor of two nodes, `p` and `q`, is defined as the lowest node in the binary tree that has both `p` and `q` as descendants.
>
> A node can also be a descendant of itself. For example, if `q` is a descendant of `p`, and we know that `p` is a descendant of itself, then `p` will be the lowest common ancestor of `p` and `q`.

**Constraints:**

* $2 \leq n \leq 500$
* All `Node.data` are unique.
* `p` $!=$ `q`
* `p` and `q` exist in the tree.

# Examples

# Understand the problem

Let’s take a moment to make sure you’ve correctly understood the problem. The quiz below helps you check if you’re solving the correct problem:

What is the lowest common ancestor of $8$ and $6$ in the following binary tree?
```text
  5
 / \
6   7
   / \
  8   9
```

What is the lowest common ancestor of <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>8</mn></mrow><annotation encoding="application/x-tex">8</annotation></semantics></math>8 and <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>6</mn></mrow><annotation encoding="application/x-tex">6</annotation></semantics></math>6 in the following binary tree?
<pre><code class="language-text"> 5
 / \
6 7
 / \
 8 9
</code></pre>

What is the lowest common ancestor of $9$ and $3$ in the following binary tree?
```text
     6 
   /   \
  9     4
 /       \
3         8
```

What is the lowest common ancestor of <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>9</mn></mrow><annotation encoding="application/x-tex">9</annotation></semantics></math>9 and <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>3</mn></mrow><annotation encoding="application/x-tex">3</annotation></semantics></math>3 in the following binary tree?
<pre><code class="language-text"> 6 
 / \
 9 4
 / \
3 8
</code></pre>

What is the lowest common ancestor of $4$ and $7$ in the following binary tree?
```text
     5 
   /   \
  2     3
 /       \
4         7
```

What is the lowest common ancestor of <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>4</mn></mrow><annotation encoding="application/x-tex">4</annotation></semantics></math>4 and <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>7</mn></mrow><annotation encoding="application/x-tex">7</annotation></semantics></math>7 in the following binary tree?
<pre><code class="language-text"> 5 
 / \
 2 3
 / \
4 7
</code></pre>

Lowest Common Ancestor in a Binary Tree

Lowest Common Ancestor in a Binary Tree

# Figure it out!

We have a game for you to play. Rearrange the logical building blocks to develop a clearer understanding of how to solve this problem.

# Try it yourself

Implement your solution in the following coding playground.

// Definiton of a binary tree node class
// class TreeNode<T> {
// T data;
// TreeNode<T> left;
// TreeNode<T> right;

// TreeNode(T data) {
// this.data = data;
// this.left = null;
// this.right = null;
// }
// }


import ds_v1.BinaryTree.TreeNode;

public class Solution {
 public TreeNode<Integer> lowestCommonAncestor(TreeNode<Integer> root, TreeNode<Integer> p, TreeNode<Integer> q) {
 
 // Replace this placeholder return statement with your code
 return null;
 }
}

from binary_tree import *
from binary_tree_node import *

class Solution:
    def __init__(self):

        # Store the lowest common ancestor in the following variable
        self.lca = None

    def lowest_common_ancestor_in_BT(self, current_node, p, q):

        # Replace this placeholder return statement with your code
        return False

# Quick Fix\nYou are Edward, a Senior Software Engineer who specializes in coding interview patterns, data structures and algorithms, and is great at debugging issues in code.\n\n- Edward, please keep in mind that you do not have to print the data of the placeholders. Also, do not mention anything about yourself.\n\nEdward, you are given the following data in the placeholders:\n- Function name: `lowestCommonAncestor`\n- Candidate's submitted solution code: USER_CODE\n- Error: ERROR_DESCRIPTION\n- Failed test case: FAILED_TEST_CASE\n- Coding language: `Java`\n\n- Edward, your task is to understand the type of error that is occurring in the coding language provided.\n- Once you've understood the error as per the error’s description, and why the test case failed, you will suggest how the candidate can improve their solution code to fix the error.\n- Edward, you will not give the corrected code, just a hint on what the error tells, and how to fix it.\n- Edward, please limit your response strictly to 100 words. No more words than the limit.

# Review Code\nEdward, your current task is to review the `lowestCommonAncestor` function(s) and the other function(s) invoked in it/them.\n\n-Take your time and review the candidate's solution code.\n- Do not review any extra/helper files or functions not invoked in the solution function `lowestCommonAncestor`.\n- Do not review the code of data structure objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. \n- Also, if the solution code doesn’t use the coding pattern stated above, then please encourage the candidate to reattempt the problem using that pattern corectly.

# Review time complexity\n- Please provide a brief time complexity analysis of the function(s) `lowestCommonAncestor`. \n- While calculating time complexity, include the time complexity of built-in Java functions invoked by the code. \n- The answer should be to the point in approximately 60 words.

# Review space complexity\n- Please provide a brief space complexity analysis of the function(s) `lowestCommonAncestor`. \n- While calculating space complexity, include the space complexity of built-in Java functions invoked by the code. \n- Do not include the space used for the input parameters and the space used for just returning an output (not used as a scratch space). \n- The answer should be to the point in approximately 60 words.

def evaluate_results(self) -> object:
 """
 self.__inputs is the list of inputs that are passed as input to function
 self.__actual_output is a list of outputs that is produced by the learner's code
 return : self.__edu__result object
 """
 
 from collections import deque

 def find(root, value):
 q = deque([root])
 while q:
 currentNode = q.popleft()
 if currentNode:
 if currentNode.data == value:
 return currentNode
 q.append(currentNode.left)
 q.append(currentNode.right)
 if all(val == None for val in q):
 break
 return None 
 
 def lowestCommonAncestor(root, p, q):
 if root is None:
 return None
 
 left_res = lowestCommonAncestor(root.left, p, q)
 right_res = lowestCommonAncestor(root.right, p, q)
 
 if (left_res and right_res) or (root in [p, q]):
 return root
 else:
 return left_res or right_res

 def level_order_traversal(root):
 if not root:
 return []
 
 result = []
 q = deque([root])
 while q:
 curr = q.popleft()
 
 if curr:
 result.append(curr.data)
 q.append(curr.left)
 q.append(curr.right)
 else:
 result.append(None)

 if all(val == None for val in q):
 break 
 
 return result
 

 def create_binary_tree(level_order):
 if not level_order:
 return None
 
 root = TreeNode(level_order[0])
 q = deque([root])
 i = 1
 while i < len(level_order) and q:
 curr = q.popleft()
 
 if level_order[i] != None:
 curr.left = TreeNode(level_order[i])
 q.append(curr.left)
 i += 1
 
 if i < len(level_order) and level_order[i] != None:
 curr.right = TreeNode(level_order[i])
 q.append(curr.right)
 i += 1
 
 return root
 

 # multiple outputs will be added at each index respectively for a single test_case
 expected_outputs = [] 
 inp = self.__inputs[0]
 p = self.__inputs[1]
 q = self.__inputs[2]
 inp_root = create_binary_tree(inp)
 expected = lowestCommonAncestor(inp_root, find(inp_root,p), find(inp_root,q))
 expected1 = expected.data
 expected_outputs.append(expected1)
 actual1 = self.__actual_outputs[0]
 
 if json.dumps(expected1) == json.dumps(actual1):
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
 else:
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

 return self.__edu_result

class TreeNode:
 def __init__(self, data):
 self.data = data
 self.left = None
 self.right = None

import ds_v1.BinaryTree.TreeNode;

class Main {
 public static int myMain(TreeNode<Integer> root, int p, int q) {
 Solution solution = new Solution();
 TreeNode<Integer> lca = solution.lowestCommonAncestor(root, find(root, p), find(root, q));
 
 if (lca!=null)
 return lca.data;

 return -1;
 }

 private static TreeNode<Integer> find(TreeNode<Integer> root, int value) {
 if (root == null || root.data.equals(value))
 return root;

 TreeNode<Integer> leftResult = find(root.left, value);
 if (leftResult != null)
 return leftResult;

 TreeNode<Integer> rightResult = find(root.right, value);
 if (rightResult != null)
 return rightResult;

 return null;
 }
}

lowest_common_ancestor.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : [1, 2, 3, 4, 5, 6]
    - 2 : 1
    - 3 : 2
    output:
    - 1 : 1
  - name: "testcase 2"
    inputs:
    - 1 : [4, 3, 6, 2, 1, 7, 5]
    - 2 : 5
    - 3 : 7
    output:
    - 1 : 6
  - name: "testcase 3"
    inputs:
    - 1 : [12, 7, 1, 9, 10, 15]
    - 2 : 1
    - 3 : 10
    output:
    - 1 : 12
  - name: "testcase 4"
    inputs:
    - 1 : [3, 9, 20, 15, 7]
    - 2 : 3
    - 3 : 9
    output:
    - 1 : 3
  - name: "testcase 5"
    inputs:
    - 1 : [1, 2]
    - 2 : 1
    - 3 : 2
    output:
    - 1 : 1

def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 self.__inputs is the list of inputs that are passed as input to function
 store the result using the self.store_result(error_msg, status)
 return : void
 """
 
 from collections import deque

 def level_order_traversal(root):
 if not root:
 return []
 
 result = []
 q = deque([root])
 while q:
 curr = q.popleft()
 
 if curr:
 result.append(curr.data)
 q.append(curr.left)
 q.append(curr.right)
 else:
 result.append(None)

 if all(val == None for val in q):
 break 
 
 return result
 
 def is_valid_level_order(level_order):
 if not level_order:
 return True # An empty list is a valid level-order traversal

 n = len(level_order)
 i = 0
 root = TreeNode(level_order[i])
 q = deque([root])
 i = 1

 while i < n and q:
 curr = q.popleft()

 if level_order[i] is not None:
 curr.left = TreeNode(level_order[i])
 q.append(curr.left)
 i += 1

 if i < n and level_order[i] is not None:
 curr.right = TreeNode(level_order[i])
 q.append(curr.right)
 i += 1
 
 # Remove None values after the last integer
 while level_order and level_order[-1] is None:
 level_order.pop()
 
 return level_order_traversal(root) == level_order

 def has_duplicates(arr):
 seen = set()
 for element in arr:
 if element is not None:
 if element in seen:
 return True
 seen.add(element)
 return False
 
 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 input_list = self.__inputs[0]
 p = self.__inputs[1]
 q = self.__inputs[2]
 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 
 if not (2 <= sum(1 for value in input_list if value is not None) <= 500):
 self.store_result ("Total nodes in the tree are expected to be between 2 and 500", ValidationFunction.FAIL)
 elif has_duplicates(input_list):
 self.store_result ("There should not duplicates in the tree", ValidationFunction.FAIL) 
 elif (len(input_list) !=0 and input_list[0] is None):
 self.store_result ("Not a valid value of type TreeNode", ValidationFunction.FAIL) 
 elif (len(input_list) !=0 and input_list[0] is None):
 self.store_result ("Not a valid value of type TreeNode", ValidationFunction.FAIL)
 elif not is_valid_level_order(input_list):
 self.store_result ("Not a valid value of type TreeNode", ValidationFunction.FAIL)
 elif not p in input_list or not q in input_list:
 self.store_result ("'p' and 'q' must exist in the tree", ValidationFunction.FAIL)
 elif p == q:
 self.store_result ("p and q are equal", ValidationFunction.FAIL)

 else:
 self.store_result("", ValidationFunction.PASS)


class TreeNode:
 def __init__(self, data):
 self.data = data
 self.left = None
 self.right = None

Try to solve the Lowest Common Ancestor of a Binary Tree problem.