Given a binary tree, how do you remove all the half nodes?
Given A binary Tree, how do you remove all the half nodes (which has only one child)? Note leaves should not be touched as they have both children as NULL.
For example consider the below tree.
Nodes 2 and 4 are half nodes as one of their child is Null.
The idea is to use post-order traversal to solve this problem efficiently. We first process the left children, then right children, and finally the node itself. So we form the new tree bottom up, starting from the leaves towards the root. By the time we process the current node, both its left and right subtrees were already processed. Below is the implementation of this idea.
C
// C program to remove all half nodes #include <stdio.h> #include <stdlib.h> struct node { int data; struct node* left, *right; }; struct node* newNode(int data) { struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = node->right = NULL; return(node); } void printInoder(struct node*root) { if (root != NULL) { printInoder(root->left); printf("%d ",root->data); printInoder(root->right); } } // Removes all nodes with only one child and returns // new root (note that root may change) struct node* RemoveHalfNodes(struct node* root) { if (root==NULL) return NULL; root->left = RemoveHalfNodes(root->left); root->right = RemoveHalfNodes(root->right); if (root->left==NULL && root->right==NULL) return root; /* if current nodes is a half node with left child NULL left, then it's right child is returned and replaces it in the given tree */ if (root->left==NULL) { struct node *new_root = root->right; free(root); // To avoid memory leak return new_root; } /* if current nodes is a half node with right child NULL right, then it's right child is returned and replaces it in the given tree */ if (root->right==NULL) { struct node *new_root = root->left; free(root); // To avoid memory leak return new_root; } return root; } // Driver program int main(void) { struct node*NewRoot=NULL; struct node *root = newNode(2); root->left = newNode(7); root->right = newNode(5); root->left->right = newNode(6); root->left->right->left=newNode(1); root->left->right->right=newNode(11); root->right->right=newNode(9); root->right->right->left=newNode(4); printf("Inorder traversal of given tree \n"); printInoder(root); NewRoot = RemoveHalfNodes(root); printf("\nInorder traversal of the modified tree \n"); printInoder(NewRoot); return 0; } |
Java
// Java program to remove half nodes class Node { int data; Node left, right; Node(int item) { data = item; left = right = null; } } class BinaryTree { Node root; void printInorder(Node node) { if (node != null) { printInorder(node.left); System.out.print(node.data + " "); printInorder(node.right); } } // Removes all nodes with only one child and returns // new root (note that root may change) Node RemoveHalfNodes(Node node) { if (node == null) return null; node.left = RemoveHalfNodes(node.left); node.right = RemoveHalfNodes(node.right); if (node.left == null && node.right == null) return node; /* if current nodes is a half node with left child NULL left, then it's right child is returned and replaces it in the given tree */ if (node.left == null) { Node new_root = node.right; return new_root; } /* if current nodes is a half node with right child NULL right, then it's right child is returned and replaces it in the given tree */ if (node.right == null) { Node new_root = node.left; return new_root; } return node; } // Driver program public static void main(String args[]) { BinaryTree tree = new BinaryTree(); Node NewRoot = null; tree.root = new Node(2); tree.root.left = new Node(7); tree.root.right = new Node(5); tree.root.left.right = new Node(6); tree.root.left.right.left = new Node(1); tree.root.left.right.right = new Node(11); tree.root.right.right = new Node(9); tree.root.right.right.left = new Node(4); System.out.println("the inorder traversal of tree is "); tree.printInorder(tree.root); NewRoot = tree.RemoveHalfNodes(tree.root); System.out.print("\nInorder traversal of the modified tree \n"); tree.printInorder(NewRoot); } } // This code has been contributed by Mayank Jaiswal |
Python
# Python program to remove all half nodes # A binary tree node class Node: # Constructor for creating a new node def __init__(self , data): self.data = data self.left = None self.right = None # For inorder traversal def printInorder(root): if root is not None: printInorder(root.left) print root.data, printInorder(root.right) # Removes all nodes with only one child and returns # new root(note that root may change) def RemoveHalfNodes(root): if root is None: return None # Recur to left tree root.left = RemoveHalfNodes(root.left) # Recur to right tree root.right = RemoveHalfNodes(root.right) # if both left and right child is None # the node is not a Half node if root.left is None and root.right is None: return root # If current nodes is a half node with left child # None then it's right child is returned and # replaces it in the given tree if root.left is None: new_root = root.right temp = root root = None del(temp) return new_root if root.right is None: new_root = root.left temp = root root = None del(temp) return new_root return root # Driver Program root = Node(2) root.left = Node(7) root.right = Node(5) root.left.right = Node(6) root.left.right.left = Node(1) root.left.right.right = Node(11) root.right.right = Node(9) root.right.right.left = Node(4) print "Inorder traversal of given tree"printInorder(root) NewRoot = RemoveHalfNodes(root) print "\nInorder traversal of the modified tree"printInorder(NewRoot) # This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System; // C# program to remove half nodes public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; } } public class BinaryTree { public Node root; public virtual void printInorder(Node node) { if (node != null) { printInorder(node.left); Console.Write(node.data + " "); printInorder(node.right); } } // Removes all nodes with only one child and returns // new root (note that root may change) public virtual Node RemoveHalfNodes(Node node) { if (node == null) { return null; } node.left = RemoveHalfNodes(node.left); node.right = RemoveHalfNodes(node.right); if (node.left == null && node.right == null) { return node; } /* if current nodes is a half node with left child NULL left, then it's right child is returned and replaces it in the given tree */ if (node.left == null) { Node new_root = node.right; return new_root; } /* if current nodes is a half node with right child NULL right, then it's right child is returned and replaces it in the given tree */ if (node.right == null) { Node new_root = node.left; return new_root; } return node; } // Driver program public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); Node NewRoot = null; tree.root = new Node(2); tree.root.left = new Node(7); tree.root.right = new Node(5); tree.root.left.right = new Node(6); tree.root.left.right.left = new Node(1); tree.root.left.right.right = new Node(11); tree.root.right.right = new Node(9); tree.root.right.right.left = new Node(4); Console.WriteLine("the inorder traversal of tree is "); tree.printInorder(tree.root); NewRoot = tree.RemoveHalfNodes(tree.root); Console.Write("\nInorder traversal of the modified tree \n"); tree.printInorder(NewRoot); } } // This code is contributed by Shrikant13 |
Output:
Inorder traversal of given tree 7 1 6 11 2 5 4 9 Inorder traversal of the modified tree 1 6 11 2 4
Time complexity of the above solution is O(n) as it does a simple traversal of binary tree.
This article is contributed by Jyoti Saini. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
