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Delete Leaf Nodes with Value as X in C++ Program
In this tutorial, we are going to learn how to delete the leaf nodes from a tree with the given value.
Let's see the steps to solve the problem.
Write a struct Node for a binary tree.
Write a function to traverse (inorder, preorder, postorder) through the tree and print all data.
Initialize the tree by creating nodes with the struct.
Initialize the x value.
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Write a function to delete the leaf nodes with the given value. It accepts two arguments root node and x value.
If the root is a null return.
Replace the left node of the root with a new root after deletion.
Same with the right node of the root.
If the current root node data is equal to x and it is a leaf node, then return a null pointer.
Return root node
Example
Let's see the code.
#include <bits/stdc++.h> using namespace std; struct Node { int data; struct Node *left, *right; }; struct Node* newNode(int data) { struct Node* newNode = new Node; newNode->data = data; newNode->left = newNode->right = NULL; return newNode; } Node* deleteLeafNodes(Node* root, int x) { if (root == NULL) { return nullptr; } root->left = deleteLeafNodes(root->left, x); root->right = deleteLeafNodes(root->right, x); // checking the current node data with x if (root->data == x && root->left == NULL && root->right == NULL) { // deleting the node return nullptr; } return root; } void inorder(Node* root) { if (root == NULL) { return; } inorder(root->left); cout << root->data << " "; inorder(root->right); } int main(void) { struct Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(3); root->left->right = newNode(4); root->right->right = newNode(5); root->right->left = newNode(4); root->right->right->left = newNode(4); root->right->right->right = newNode(4); deleteLeafNodes(root, 4); cout << "Tree: "; inorder(root); cout << endl; return 0; }
Output
If you execute the above code, then you will get the following result.
Tree: 3 2 1 3 5
Conclusion
If you have any queries in the tutorial, mention them in the comment section.