Given a binary tree and a target integer x, delete all the leaf nodes having value as x. Also, delete the newly formed leaves with the target value as x.
Example:
Input : x = 5
6
/ \
5 4
/ \ \
1 2 5
Output :
6
/ \
5 4
/ \
1 2
Inorder Traversal is 1 5 2 6 4
Source: Microsoft Interview
We traverse the tree in postorder fashion and recursively delete the nodes. The approach is very similar to this and this problem.
// CPP code to delete all leaves with given
// value.
#include <bits/stdc++.h>
using namespace std;
// A binary tree node
struct Node {
int data;
struct Node *left, *right;
};
// A utility function to allocate a new node
struct Node* newNode(int data)
{
struct Node* newNode = new Node;
newNode->data = data;
newNode->left = newNode->right = NULL;
return (newNode);
}
Node* deleteLeaves(Node* root, int x)
{
if (root == NULL)
return nullptr;
root->left = deleteLeaves(root->left, x);
root->right = deleteLeaves(root->right, x);
if (root->data == x && root->left == NULL &&
root->right == NULL) {
return nullptr;
}
return root;
}
void inorder(Node* root)
{
if (root == NULL)
return;
inorder(root->left);
cout << root->data << " ";
inorder(root->right);
}
// Driver program
int main(void)
{
struct Node* root = newNode(10);
root->left = newNode(3);
root->right = newNode(10);
root->left->left = newNode(3);
root->left->right = newNode(1);
root->right->right = newNode(3);
root->right->right->left = newNode(3);
root->right->right->right = newNode(3);
deleteLeaves(root, 3);
cout << "Inorder traversal after deletion : ";
inorder(root);
return 0;
}
// Java code to delete all leaves with given
// value.
class GfG {
// A binary tree node
static class Node {
int data;
Node left, right;
}
// A utility function to allocate a new node
static Node newNode(int data)
{
Node newNode = new Node();
newNode.data = data;
newNode.left = null;
newNode.right = null;
return (newNode);
}
static Node deleteLeaves(Node root, int x)
{
if (root == null)
return null;
root.left = deleteLeaves(root.left, x);
root.right = deleteLeaves(root.right, x);
if (root.data == x && root.left == null && root.right == null) {
return null;
}
return root;
}
static void inorder(Node root)
{
if (root == null)
return;
inorder(root.left);
System.out.print(root.data + " ");
inorder(root.right);
}
// Driver program
public static void main(String[] args)
{
Node root = newNode(10);
root.left = newNode(3);
root.right = newNode(10);
root.left.left = newNode(3);
root.left.right = newNode(1);
root.right.right = newNode(3);
root.right.right.left = newNode(3);
root.right.right.right = newNode(3);
deleteLeaves(root, 3);
System.out.print("Inorder traversal after deletion : ");
inorder(root);
}
}
# Python3 code to delete all leaves
# with given value.
# A utility class to allocate a new node
class newNode:
def __init__(self, data):
self.data = data
self.left = self.right = None
def deleteLeaves(root, x):
if (root == None):
return None
root.left = deleteLeaves(root.left, x)
root.right = deleteLeaves(root.right, x)
if (root.data == x and
root.left == None and
root.right == None):
return None
return root
def inorder(root):
if (root == None):
return
inorder(root.left)
print(root.data, end = " ")
inorder(root.right)
# Driver Code
if __name__ == '__main__':
root = newNode(10)
root.left = newNode(3)
root.right = newNode(10)
root.left.left = newNode(3)
root.left.right = newNode(1)
root.right.right = newNode(3)
root.right.right.left = newNode(3)
root.right.right.right = newNode(3)
deleteLeaves(root, 3)
print("Inorder traversal after deletion : ")
inorder(root)
# This code is contributed by PranchalK
// C# code to delete all leaves with given
// value.
using System;
class GfG
{
// A binary tree node
class Node
{
public int data;
public Node left, right;
}
// A utility function to allocate a new node
static Node newNode(int data)
{
Node newNode = new Node();
newNode.data = data;
newNode.left = null;
newNode.right = null;
return (newNode);
}
static Node deleteLeaves(Node root, int x)
{
if (root == null)
return null;
root.left = deleteLeaves(root.left, x);
root.right = deleteLeaves(root.right, x);
if (root.data == x &&
root.left == null &&
root.right == null)
{
return null;
}
return root;
}
static void inorder(Node root)
{
if (root == null)
return;
inorder(root.left);
Console.Write(root.data + " ");
inorder(root.right);
}
// Driver code
public static void Main(String[] args)
{
Node root = newNode(10);
root.left = newNode(3);
root.right = newNode(10);
root.left.left = newNode(3);
root.left.right = newNode(1);
root.right.right = newNode(3);
root.right.right.left = newNode(3);
root.right.right.right = newNode(3);
deleteLeaves(root, 3);
Console.Write("Inorder traversal after deletion : ");
inorder(root);
}
}
// This code is contributed by PrinciRaj1992
<script>
// JavaScript code to delete all leaves with given value.
class Node
{
constructor(data) {
this.left = null;
this.right = null;
this.data = data;
}
}
// A utility function to allocate a new node
function newNode(data)
{
let newNode = new Node(data);
return (newNode);
}
function deleteLeaves(root, x)
{
if (root == null)
return null;
root.left = deleteLeaves(root.left, x);
root.right = deleteLeaves(root.right, x);
if (root.data == x && root.left == null && root.right == null)
{
return null;
}
return root;
}
function inorder(root)
{
if (root == null)
return;
inorder(root.left);
document.write(root.data + " ");
inorder(root.right);
}
let root = newNode(10);
root.left = newNode(3);
root.right = newNode(10);
root.left.left = newNode(3);
root.left.right = newNode(1);
root.right.right = newNode(3);
root.right.right.left = newNode(3);
root.right.right.right = newNode(3);
deleteLeaves(root, 3);
document.write("Inorder traversal after deletion : ");
inorder(root);
</script>
Output:
Inorder traversal after deletion : 3 1 10 10
Time complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(h) where h is the height of the binary tree.