Maximum in Binary tree

Abhishek Kumar • Category: Data Structures • Tags: datastructures binary-tree

Find the maximum element in a Binary tree using recursion or iterative solution. For example, a binary tree contains the following elements-
1, 2, 3, 4, 5, 6, 7
Here the maximum element will be 7.

In a Binary Tree, to find the maximum element we have to traverse each element and check if an element higher than the current one is available then assign it as the maximum element. In the same way, we can find the minimum element too.

We can solve this problem in two ways-

Recursive solution
Iterative solution

Recursive solution

For recursive solution we have to check each node one by one recursively as we have done in binary tree recursive traversal.

Python
Javascript

class Node:
    def __init__(self, data):
        self.left = None
        self.data = data
        self.right = None

def find_max(root):
    if root is None:
        return float('-inf')
    max_val = root.data
    left_val = find_max(root.left)
    right_val = find_max(root.right)
    if left_val > max_val:
        max_val = left_val
    if right_val > max_val:
        max_val = right_val
    return max_val

root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
print(find_max(root))

class Node {
  constructor(data) {
  this.left = null;
  this.data = data;
  this.right = null;
  }
}

function findMax(root) {
  if (root == null) {
    return Number.NEGATIVE_INFINITY;
  }
  let maxVal = root.data;
  leftVal = findMax(root.left);
  rightVal = findMax(root.right);
  if (leftVal > maxVal) {
    maxVal = leftVal;
  }
  if (rightVal > maxVal) {
    maxVal = rightVal;
  }
  return maxVal;
}

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
console.log(findMax(root));

Output

Iterative solution

For iterative solution, we have to traverse each node in level order and find the maximum by comparision.

Python
Javascript

import queue

class Node:
    def __init__(self, data):
        self.left = None
        self.data = data
        self.right = None

def find_max(root):
    if root:
        max_val = root.data
        q = queue.Queue()
        q.put(root)
        while not q.empty():
            node = q.get()
            if node.data > max_val:
                max_val = node.data
            if node.left:
                q.put(node.left)
            if node.right:
                q.put(node.right)
        return max_val
    else:
        return None

root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
print(find_max(root))

class Node {
  constructor(data) {
    this.left = null;
    this.data = data;
    this.right = null;
  }
}

function findMax(root) {
  if (root) {
    const q = [];
    q.push(root);
    node = null;

    let maxVal = root.data;
    while (q.length > 0) {
      node = q.shift();

      if (node.data > maxVal) {
        maxVal = node.data;
      }
      if (node.left) {
        q.push(node.left);
      }
      if (node.right) {
        q.push(node.right);
      }
    }
    return maxVal;
  } else {
    return null;
  }
}

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
console.log(findMax(root));