Delete the last leaf node in a Binary Tree
Given a Binary Tree, the task is to find and DELETE the last leaf node.
The leaf node is a node with no children. The last leaf node would be the node that is traversed last in sequence during Level Order Traversal. The problem statement is to identify this last visited node and delete this particular node.
Examples:
Input:
Given Tree is:
6
/ \
5 4
/ \ \
1 2 5
Level Order Traversal is: 6 5 4 1 2 5
Output:
After deleting the last node (5),
the tree would look like as follows.
6
/ \
5 4
/ \
1 2
Level Order Traversal is: 6 5 4 1 2
Input:
Given tree is:
1
/ \
3 10
/ \ / \
2 15 4 5
/
1
Level Order Traversal is: 1 3 10 2 15 4 5 1
Output:
After deleting the last node (1),
the tree would look like as follows.
1
/ \
3 10
/ \ / \
2 15 4 5
Level Order Traversal is: 1 3 10 2 15 4 5
This problem is slightly different from Delete leaf node with value as X wherein we are right away given the value of last leaf node (X) to be deleted, based on which we perform checks and mark the parent node as null to delete it.
This approach would identify the last present leaf node on last level of the tree and would delete it.
Approach 1: Traversing last level nodes and keeping track of Parent and traversed node.
This approach would traverse each node until we reach the last level of the given binary tree. While traversing, we keep track of the last traversed node and its Parent.
Once done with traversal, Check if the parent has Right Child, if yes, set it to NULL. If no, set the left pointer to NULL
Below is the implementation of the approach:
C++
// CPP implementation of the approach#include <bits/stdc++.h>using namespace std; // Tree Nodeclass Node{public: int data; Node *left, *right; Node(int data) : data(data) {}}; // Method to perform inorder traversalvoid inorder(Node *root){ if (root == NULL) return; inorder(root->left); cout << root->data << " "; inorder(root->right);} // To keep track of last processed// nodes parent and node itself.Node *lastNode, *parentOfLastNode; // Method to get the height of the treeint height(Node *root){ if (root == NULL) return 0; int lheight = height(root->left) + 1; int rheight = height(root->right) + 1; return max(lheight, rheight);} // Method to keep track of parents// of every nodevoid getLastNodeAndItsParent(Node *root, int level, Node *parent){ if (root == NULL) return; // The last processed node in // Level Order Traversal has to // be the node to be deleted. // This will store the last // processed node and its parent. if (level == 1) { lastNode = root; parentOfLastNode = parent; } getLastNodeAndItsParent(root->left, level - 1, root); getLastNodeAndItsParent(root->right, level - 1, root);} // Method to delete last leaf nodevoid deleteLastNode(Node *root){ int levelOfLastNode = height(root); getLastNodeAndItsParent(root, levelOfLastNode, NULL); if (lastNode and parentOfLastNode) { if (parentOfLastNode->right) parentOfLastNode->right = NULL; else parentOfLastNode->left = NULL; } else cout << "Empty Tree\n";} // Driver Codeint main(){ Node *root = new Node(6); root->left = new Node(5); root->right = new Node(4); root->left->left = new Node(1); root->left->right = new Node(2); root->right->right = new Node(5); cout << "Inorder traversal before deletion of last node :\n"; inorder(root); deleteLastNode(root); cout << "\nInorder traversal after deletion of last node :\n"; inorder(root); return 0;} // This code is contributed by// sanjeev2552 |
Java
// Java Implementation of the approachpublic class DeleteLastNode { // Tree Node static class Node { Node left, right; int data; Node(int data) { this.data = data; } } // Method to perform inorder traversal public void inorder(Node root) { if (root == null) return; inorder(root.left); System.out.print(root.data + " "); inorder(root.right); } // To keep track of last processed // nodes parent and node itself. public static Node lastNode; public static Node parentOfLastNode; // Method to get the height of the tree public int height(Node root) { if (root == null) return 0; int lheight = height(root.left) + 1; int rheight = height(root.right) + 1; return Math.max(lheight, rheight); } // Method to delete last leaf node public void deleteLastNode(Node root) { int levelOfLastNode = height(root); // Get all nodes at last level getLastNodeAndItsParent(root, levelOfLastNode, null); if (lastNode != null && parentOfLastNode != null) { if (parentOfLastNode.right != null) parentOfLastNode.right = null; else parentOfLastNode.left = null; } else System.out.println("Empty Tree"); } // Method to keep track of parents // of every node public void getLastNodeAndItsParent(Node root, int level, Node parent) { if (root == null) return; // The last processed node in // Level Order Traversal has to // be the node to be deleted. // This will store the last // processed node and its parent. if (level == 1) { lastNode = root; parentOfLastNode = parent; } getLastNodeAndItsParent(root.left, level - 1, root); getLastNodeAndItsParent(root.right, level - 1, root); } // Driver Code public static void main(String[] args) { Node root = new Node(6); root.left = new Node(5); root.right = new Node(4); root.left.left = new Node(1); root.left.right = new Node(2); root.right.right = new Node(5); DeleteLastNode deleteLastNode = new DeleteLastNode(); System.out.println("Inorder traversal " + "before deletion " + "of last node : "); deleteLastNode.inorder(root); deleteLastNode.deleteLastNode(root); System.out.println("\nInorder traversal " + "after deletion of " + "last node : "); deleteLastNode.inorder(root); }} |
C#
// C# implementation of the above approachusing System; class GFG{ // Tree Node public class Node { public Node left, right; public int data; public Node(int data) { this.data = data; } } // Method to perform inorder traversal public void inorder(Node root) { if (root == null) return; inorder(root.left); Console.Write(root.data + " "); inorder(root.right); } // To keep track of last processed // nodes parent and node itself. public static Node lastNode; public static Node parentOfLastNode; // Method to get the height of the tree public int height(Node root) { if (root == null) return 0; int lheight = height(root.left) + 1; int rheight = height(root.right) + 1; return Math.Max(lheight, rheight); } // Method to delete last leaf node public void deleteLastNode(Node root) { int levelOfLastNode = height(root); // Get all nodes at last level getLastNodeAndItsParent(root, levelOfLastNode, null); if (lastNode != null && parentOfLastNode != null) { if (parentOfLastNode.right != null) parentOfLastNode.right = null; else parentOfLastNode.left = null; } else Console.WriteLine("Empty Tree"); } // Method to keep track of parents // of every node public void getLastNodeAndItsParent(Node root, int level, Node parent) { if (root == null) return; // The last processed node in // Level Order Traversal has to // be the node to be deleted. // This will store the last // processed node and its parent. if (level == 1) { lastNode = root; parentOfLastNode = parent; } getLastNodeAndItsParent(root.left, level - 1, root); getLastNodeAndItsParent(root.right, level - 1, root); } // Driver Code public static void Main(String[] args) { Node root = new Node(6); root.left = new Node(5); root.right = new Node(4); root.left.left = new Node(1); root.left.right = new Node(2); root.right.right = new Node(5); GFG deleteLastNode = new GFG(); Console.WriteLine("Inorder traversal " + "before deletion " + "of last node : "); deleteLastNode.inorder(root); deleteLastNode.deleteLastNode(root); Console.WriteLine("\nInorder traversal " + "after deletion of " + "last node : "); deleteLastNode.inorder(root); }} // This code is contributed by 29AjayKumar |
Output:
Inorder traversal before deletion of last node : 1 5 2 6 4 5 Inorder traversal after deletion of last node : 1 5 2 6 4
Time Complexity: 
Since each node would be traversed once, the time taken would be linear to the number of nodes in a given tree.
Approach 2: Performing Level Order Traversal on given Binary Tree using Queue and tracking Parent and last traversed node.
This is a Non-Recursive way of achieving above Approach 1. We perform the Level Order Traversal using Queue and keeping track of every visited node and its parent. The last visited node would be the last node that is to be deleted.
Below is the implementation of the approach:
Java
// Java implementationimport java.util.LinkedList;import java.util.Queue; public class DeleteLastNode { // Tree Node static class Node { Node left, right; int data; Node(int data) { this.data = data; } } // Function to perform the inorder traversal of the tree public void inorder(Node root) { if (root == null) return; inorder(root.left); System.out.print(root.data + " "); inorder(root.right); } // To keep track of last // processed nodes parent // and node itself. public static Node lastLevelLevelOrder; public static Node parentOfLastNode; // Method to delete the last node // from the tree public void deleteLastNode(Node root) { // If tree is empty, it // would return without // any deletion if (root == null) return; // The queue would be needed // to maintain the level order // traversal of nodes Queue<Node> queue = new LinkedList<>(); queue.offer(root); // The traversing would // continue untill all // nodes are traversed once while (!queue.isEmpty()) { Node temp = queue.poll(); // If there is left child if (temp.left != null) { queue.offer(temp.left); // For every traverssed node, // we would check if it is a // leaf node by checking if // current node has children to it if (temp.left.left == null && temp.left.right == null) { // For every leaf node // encountered, we would // keep not of it as // "Previously Visided Leaf node. lastLevelLevelOrder = temp.left; parentOfLastNode = temp; } } if (temp.right != null) { queue.offer(temp.right); if (temp.right.left == null && temp.right.right == null) { // For every leaf node // encountered, we would // keep not of it as // "Previously Visided Leaf node. lastLevelLevelOrder = temp.right; parentOfLastNode = temp; } } } // Once out of above loop. // we would certainly have // last visited node, which // is to be deleted and its // parent node. if (lastLevelLevelOrder != null && parentOfLastNode != null) { // If last node is right child // of parent, make right node // of its parent as NULL or // make left node as NULL if (parentOfLastNode.right != null) parentOfLastNode.right = null; else parentOfLastNode.left = null; } else System.out.println("Empty Tree"); } // Driver Code public static void main(String[] args) { Node root = new Node(6); root.left = new Node(5); root.right = new Node(4); root.left.left = new Node(1); root.left.right = new Node(2); root.right.right = new Node(5); DeleteLastNode deleteLastNode = new DeleteLastNode(); System.out.println("Inorder traversal " + "before deletion of " + "last node : "); deleteLastNode.inorder(root); deleteLastNode.deleteLastNode(root); System.out.println("\nInorder traversal " + "after deletion " + "of last node : "); deleteLastNode.inorder(root); }} |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;public class DeleteLastNode { // Tree Node public class Node { public Node left, right; public int data; public Node(int data) { this.data = data; } } // Function to perform the inorder traversal of the tree public void inorder(Node root) { if (root == null) return; inorder(root.left); Console.Write(root.data + " "); inorder(root.right); } // To keep track of last // processed nodes parent // and node itself. public static Node lastLevelLevelOrder; public static Node parentOfLastNode; // Method to delete the last node // from the tree public void deleteLastNode(Node root) { // If tree is empty, it // would return without // any deletion if (root == null) return; // The queue would be needed // to maintain the level order // traversal of nodes Queue<Node> queue = new Queue<Node>(); queue.Enqueue(root); // The traversing would // continue untill all // nodes are traversed once while (queue.Count!=0) { Node temp = queue.Dequeue(); // If there is left child if (temp.left != null) { queue.Enqueue(temp.left); // For every traverssed node, // we would check if it is a // leaf node by checking if // current node has children to it if (temp.left.left == null && temp.left.right == null) { // For every leaf node // encountered, we would // keep not of it as // "Previously Visided Leaf node. lastLevelLevelOrder = temp.left; parentOfLastNode = temp; } } if (temp.right != null) { queue.Enqueue(temp.right); if (temp.right.left == null && temp.right.right == null) { // For every leaf node // encountered, we would // keep not of it as // "Previously Visided Leaf node. lastLevelLevelOrder = temp.right; parentOfLastNode = temp; } } } // Once out of above loop. // we would certainly have // last visited node, which // is to be deleted and its // parent node. if (lastLevelLevelOrder != null && parentOfLastNode != null) { // If last node is right child // of parent, make right node // of its parent as NULL or // make left node as NULL if (parentOfLastNode.right != null) parentOfLastNode.right = null; else parentOfLastNode.left = null; } else Console.WriteLine("Empty Tree"); } // Driver Code public static void Main(String[] args) { Node root = new Node(6); root.left = new Node(5); root.right = new Node(4); root.left.left = new Node(1); root.left.right = new Node(2); root.right.right = new Node(5); DeleteLastNode deleteLastNode = new DeleteLastNode(); Console.WriteLine("Inorder traversal " + "before deletion of " + "last node : "); deleteLastNode.inorder(root); deleteLastNode.deleteLastNode(root); Console.WriteLine("\nInorder traversal " + "after deletion " + "of last node : "); deleteLastNode.inorder(root); }} // This code contributed by Rajput-Ji |
Output:
Inorder traversal before deletion of last node : 1 5 2 6 4 5 Inorder traversal after deletion of last node : 1 5 2 6 4
Time Complexity:
Since every node would be visited once, the time taken would be linear to the number of nodes present in the tree.
Auxiliary Space:
Since we would be maintaining a queue to do the level order traversal, the space consumed would be .
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