Get maximum left node in binary tree
Given a tree, the task is to find the maximum in an only left node of the binary tree.
Examples:
Input :
7
/ \
6 5
/ \ / \
4 3 2 1
Output :
6
Input :
1
/ \
2 3
/ / \
4 5 6
\ / \
7 8 9
Output :
8
Traverse with inorder traversal and Apply the condition for the left node only and get maximum of left node.
Let’s try to understand with code.
C++
// CPP program to print maximum element // in left node. #include <bits/stdc++.h> using namespace std; // A Binary Tree Node struct Node { int data; struct Node *left, *right; }; // Get max of left element using // Inorder traversal int maxOfLeftElement(Node* root) { int res = INT_MIN; if (root == NULL) return res; if (root->left != NULL) res = root->left->data; // Return maximum of three values // 1) Recursive max in left subtree // 2) Value in left node // 3) Recursive max in right subtree return max({ maxOfLeftElement(root->left), res, maxOfLeftElement(root->right) }); } // Utility function to create a new tree node Node* newNode(int data) { Node* temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp; } // Driver program to test above functions int main() { // Let us create binary tree shown in above diagram Node* root = newNode(7); root->left = newNode(6); root->right = newNode(5); root->left->left = newNode(4); root->left->right = newNode(3); root->right->left = newNode(2); root->right->right = newNode(1); /* 7 / \ 6 5 / \ / \ 4 3 2 1 */ cout << maxOfLeftElement(root); return 0; } |
Java
// Java program to print maximum element // in left node. import java.util.*; class GfG { // A Binary Tree Node static class Node { int data; Node left, right; } // Get max of left element using // Inorder traversal static int maxOfLeftElement(Node root) { int res = Integer.MIN_VALUE; if (root == null) return res; if (root.left != null) res = root.left.data; // Return maximum of three values // 1) Recursive max in left subtree // 2) Value in left node // 3) Recursive max in right subtree return Math.max(maxOfLeftElement(root.left), Math.max(res, maxOfLeftElement(root.right))); } // Utility function to create a new tree node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.left = null; temp.right = null; return temp; } // Driver program to test above functions public static void main(String[] args) { // Let us create binary tree shown in above diagram Node root = newNode(7); root.left = newNode(6); root.right = newNode(5); root.left.left = newNode(4); root.left.right = newNode(3); root.right.left = newNode(2); root.right.right = newNode(1); /* 7 / \ 6 5 / \ / \ 4 3 2 1 */ System.out.println(maxOfLeftElement(root)); } } |
Python3
# Python program to prmaximum element # in left node. # Utility class to create a # new tree node class newNode: def __init__(self, data): self.data = data self.left = self.right = None # Get max of left element using # Inorder traversal def maxOfLeftElement(root): res = -999999999999 if (root == None): return res if (root.left != None): res = root.left.data # Return maximum of three values # 1) Recursive max in left subtree # 2) Value in left node # 3) Recursive max in right subtree return max({ maxOfLeftElement(root.left), res, maxOfLeftElement(root.right) }) # Driver Code if __name__ == '__main__': # Let us create binary tree shown # in above diagram root = newNode(7) root.left = newNode(6) root.right = newNode(5) root.left.left = newNode(4) root.left.right = newNode(3) root.right.left = newNode(2) root.right.right = newNode(1) # 7 # / \ # 6 5 # / \ / \ # 4 3 2 1 print(maxOfLeftElement(root)) # This code is contributed by PranchalK |
C#
// C# program to print maximum element // in left node. using System; class GfG { // A Binary Tree Node class Node { public int data; public Node left, right; } // Get max of left element using // Inorder traversal static int maxOfLeftElement(Node root) { int res = int.MinValue; if (root == null) return res; if (root.left != null) res = root.left.data; // Return maximum of three values // 1) Recursive max in left subtree // 2) Value in left node // 3) Recursive max in right subtree return Math.Max(maxOfLeftElement(root.left), Math.Max(res, maxOfLeftElement(root.right))); } // Utility function to create a new tree node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.left = null; temp.right = null; return temp; } // Driver code public static void Main(String[] args) { // Let us create binary tree // shown in above diagram Node root = newNode(7); root.left = newNode(6); root.right = newNode(5); root.left.left = newNode(4); root.left.right = newNode(3); root.right.left = newNode(2); root.right.right = newNode(1); /* 7 / \ 6 5 / \ / \ 4 3 2 1 */ Console.WriteLine(maxOfLeftElement(root)); } } // This code is contributed by 29AjayKumar |
Output:
6
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