Reverse a path in BST using queue
Last Updated :
17 Aug, 2022
Given a binary search tree and a key, your task to reverse path of the binary tree.
Prerequisite : Reverse path of Binary tree
Examples :
Input : 50
/ \
30 70
/ \ / \
20 40 60 80
k = 70
Output :
Inorder before reversal :
20 30 40 50 60 70 80
Inorder after reversal :
20 30 40 70 60 50 80
Input : 8
/ \
3 10
/ \ \
1 6 14
/ \ /
4 7 13
k = 13
Output :
Inorder before reversal :
1 3 4 6 7 8 10 13 14
Inorder after reversal :
1 3 4 6 7 13 14 8 10
Approach: Take a queue and push all the element till that given key at the end replace node key with queue front element till root, then print inorder of the tree.
Below is the implementation of above approach :
C++
// C++ code to demonstrate insert
// operation in binary search tree
#include <bits/stdc++.h>
using namespace std;
struct node {
int key;
struct node *left, *right;
};
// A utility function to
// create a new BST node
struct node* newNode(int item)
{
struct node* temp = new node;
temp->key = item;
temp->left = temp->right = NULL;
return temp;
}
// A utility function to
// do inorder traversal of BST
void inorder(struct node* root)
{
if (root != NULL) {
inorder(root->left);
cout << root->key << " ";
inorder(root->right);
}
}
// reverse tree path using queue
void reversePath(struct node** node,
int& key, queue<int>& q1)
{
/* If the tree is empty,
return a new node */
if (node == NULL)
return;
// If the node key equal
// to key then
if ((*node)->key == key)
{
// push current node key
q1.push((*node)->key);
// replace first node
// with last element
(*node)->key = q1.front();
// remove first element
q1.pop();
// return
return;
}
// if key smaller than node key then
else if (key < (*node)->key)
{
// push node key into queue
q1.push((*node)->key);
// recursive call itself
reversePath(&(*node)->left, key, q1);
// replace queue front to node key
(*node)->key = q1.front();
// perform pop in queue
q1.pop();
}
// if key greater than node key then
else if (key > (*node)->key)
{
// push node key into queue
q1.push((*node)->key);
// recursive call itself
reversePath(&(*node)->right, key, q1);
// replace queue front to node key
(*node)->key = q1.front();
// perform pop in queue
q1.pop();
}
// return
return;
}
/* A utility function to insert
a new node with given key in BST */
struct node* insert(struct node* node,
int key)
{
/* If the tree is empty,
return a new node */
if (node == NULL)
return newNode(key);
/* Otherwise, recur down the tree */
if (key < node->key)
node->left = insert(node->left, key);
else if (key > node->key)
node->right = insert(node->right, key);
/* return the (unchanged) node pointer */
return node;
}
// Driver Program to test above functions
int main()
{
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
struct node* root = NULL;
queue<int> q1;
// reverse path till k
int k = 80;
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
cout << "Before Reverse :" << endl;
// print inorder traversal of the BST
inorder(root);
cout << "\n";
// reverse path till k
reversePath(&root, k, q1);
cout << "After Reverse :" << endl;
// print inorder of reverse path tree
inorder(root);
return 0;
}
// Java code to demonstrate insert
// operation in binary search tree
import java.util.*;
class GFG
{
static class node
{
int key;
node left, right;
};
static node root = null;
static Queue<Integer> q1;
static int k;
// A utility function to
// create a new BST node
static node newNode(int item)
{
node temp = new node();
temp.key = item;
temp.left = temp.right = null;
return temp;
}
// A utility function to
// do inorder traversal of BST
static void inorder(node root)
{
if (root != null)
{
inorder(root.left);
System.out.print(root.key + " ");
inorder(root.right);
}
}
// reverse tree path using queue
static void reversePath(node node)
{
/* If the tree is empty,
return a new node */
if (node == null)
return;
// If the node key equal
// to key then
if ((node).key == k)
{
// push current node key
q1.add((node).key);
// replace first node
// with last element
(node).key = q1.peek();
// remove first element
q1.remove();
// return
return;
}
// if key smaller than node key then
else if (k < (node).key)
{
// push node key into queue
q1.add((node).key);
// recursive call itself
reversePath((node).left);
// replace queue front to node key
(node).key = q1.peek();
// perform pop in queue
q1.remove();
}
// if key greater than node key then
else if (k > (node).key)
{
// push node key into queue
q1.add((node).key);
// recursive call itself
reversePath((node).right);
// replace queue front to node key
(node).key = q1.peek();
// perform pop in queue
q1.remove();
}
// return
return;
}
/* A utility function to insert
a new node with given key in BST */
static node insert(node node, int key)
{
/* If the tree is empty,
return a new node */
if (node == null)
return newNode(key);
/* Otherwise, recur down the tree */
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
/* return the (unchanged) node pointer */
return node;
}
// Driver code
public static void main(String[] args)
{
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
q1 = new LinkedList<>();
// reverse path till k
k = 80;
root = insert(root, 50);
root = insert(root, 30);
root = insert(root, 20);
root = insert(root, 40);
root = insert(root, 70);
root = insert(root, 60);
root = insert(root, 80);
System.out.print("Before Reverse :"
+ "\n");
// print inorder traversal of the BST
inorder(root);
System.out.print("\n");
// reverse path till k
reversePath(root);
System.out.print("After Reverse :"
+ "\n");
// print inorder of reverse path tree
inorder(root);
}
}
// This code is contributed by gauravrajput1
# Python3 code to demonstrate insert
# operation in binary search tree
class Node:
# Constructor to create a new node
def __init__(self, data):
self.key = data
self.left = None
self.right = None
# A utility function to
# do inorder traversal of BST
def inorder(root):
if root != None:
inorder(root.left)
print(root.key, end = " ")
inorder(root.right)
# reverse tree path using queue
def reversePath(node, key, q1):
# If the tree is empty,
# return a new node */
if node == None:
return
# If the node key equal
# to key then
if node.key == key:
# push current node key
q1.insert(0, node.key)
# replace first node
# with last element
node.key = q1[-1]
# remove first element
q1.pop()
# return
return
# if key smaller than node key then
elif key < node.key:
# push node key into queue
q1.insert(0, node.key)
# recursive call itself
reversePath(node.left, key, q1)
# replace queue front to node key
node.key = q1[-1]
# perform pop in queue
q1.pop()
# if key greater than node key then
elif (key > node.key):
# push node key into queue
q1.insert(0, node.key)
# recursive call itself
reversePath(node.right, key, q1)
# replace queue front to node key
node.key = q1[-1]
# perform pop in queue
q1.pop()
# return
return
# A utility function to insert
#a new node with given key in BST */
def insert(node, key):
# If the tree is empty,
# return a new node */
if node == None:
return Node(key)
# Otherwise, recur down the tree */
if key < node.key:
node.left = insert(node.left, key)
elif key > node.key:
node.right = insert(node.right, key)
# return the (unchanged) node pointer */
return node
# Driver Code
if __name__ == '__main__':
# Let us create following BST
# 50
# / \
# 30 70
# / \ / \
# 20 40 60 80 */
root = None
q1 = []
# reverse path till k
k = 80;
root = insert(root, 50)
insert(root, 30)
insert(root, 20)
insert(root, 40)
insert(root, 70)
insert(root, 60)
insert(root, 80)
print("Before Reverse :")
# print inorder traversal of the BST
inorder(root)
# reverse path till k
reversePath(root, k, q1)
print()
print("After Reverse :")
# print inorder of reverse path tree
inorder(root)
# This code is contributed by PranchalK
// C# code to demonstrate insert
// operation in binary search tree
using System;
using System.Collections.Generic;
class GFG{
public class node
{
public int key;
public node left, right;
};
static node root = null;
static Queue<int> q1;
static int k;
// A utility function to
// create a new BST node
static node newNode(int item)
{
node temp = new node();
temp.key = item;
temp.left = temp.right = null;
return temp;
}
// A utility function to
// do inorder traversal of BST
static void inorder(node root)
{
if (root != null)
{
inorder(root.left);
Console.Write(root.key + " ");
inorder(root.right);
}
}
// Reverse tree path using queue
static void reversePath(node node)
{
// If the tree is empty,
// return a new node
if (node == null)
return;
// If the node key equal
// to key then
if ((node).key == k)
{
// push current node key
q1.Enqueue((node).key);
// replace first node
// with last element
(node).key = q1.Peek();
// Remove first element
q1.Dequeue();
// Return
return;
}
// If key smaller than node key then
else if (k < (node).key)
{
// push node key into queue
q1.Enqueue((node).key);
// Recursive call itself
reversePath((node).left);
// Replace queue front to node key
(node).key = q1.Peek();
// Perform pop in queue
q1.Dequeue();
}
// If key greater than node key then
else if (k > (node).key)
{
// push node key into queue
q1.Enqueue((node).key);
// Recursive call itself
reversePath((node).right);
// Replace queue front to node key
(node).key = q1.Peek();
// Perform pop in queue
q1.Dequeue();
}
// Return
return;
}
// A utility function to insert
// a new node with given key in BST
static node insert(node node, int key)
{
// If the tree is empty,
// return a new node
if (node == null)
return newNode(key);
// Otherwise, recur down the tree
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
// Return the (unchanged) node pointer
return node;
}
// Driver code
public static void Main(String[] args)
{
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
q1 = new Queue<int>();
// Reverse path till k
k = 80;
root = insert(root, 50);
root = insert(root, 30);
root = insert(root, 20);
root = insert(root, 40);
root = insert(root, 70);
root = insert(root, 60);
root = insert(root, 80);
Console.Write("Before Reverse :" + "\n");
// Print inorder traversal of the BST
inorder(root);
Console.Write("\n");
// Reverse path till k
reversePath(root);
Console.Write("After Reverse :" + "\n");
// Print inorder of reverse path tree
inorder(root);
}
}
// This code is contributed by gauravrajput1
<script>
// javascript code to demonstrate insert
// operation in binary search tree
class node
{
constructor()
{
this.key = 0;
this.left = null;
this.right = null;
}
};
var root = null;
var q1 = [];
var k = 0;
// A utility function to
// create a new BST node
function newNode(item)
{
var temp = new node();
temp.key = item;
temp.left = temp.right = null;
return temp;
}
// A utility function to
// do inorder traversal of BST
function inorder(root)
{
if (root != null)
{
inorder(root.left);
document.write(root.key + " ");
inorder(root.right);
}
}
// Reverse tree path using queue
function reversePath(node)
{
// If the tree is empty,
// return a new node
if (node == null)
return;
// If the node key equal
// to key then
if ((node).key == k)
{
// push current node key
q1.push((node).key);
// replace first node
// with last element
(node).key = q1[0];
// Remove first element
q1.shift();
// Return
return;
}
// If key smaller than node key then
else if (k < (node).key)
{
// push node key into queue
q1.push((node).key);
// Recursive call itself
reversePath((node).left);
// Replace queue front to node key
(node).key = q1[0];
// Perform pop in queue
q1.shift();
}
// If key greater than node key then
else if (k > (node).key)
{
// push node key into queue
q1.push((node).key);
// Recursive call itself
reversePath((node).right);
// Replace queue front to node key
(node).key = q1[0];
// Perform pop in queue
q1.shift();
}
// Return
return;
}
// A utility function to insert
// a new node with given key in BST
function insert(node, key)
{
// If the tree is empty,
// return a new node
if (node == null)
return newNode(key);
// Otherwise, recur down the tree
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
// Return the (unchanged) node pointer
return node;
}
// Driver code
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
q1 = [];
// Reverse path till k
k = 80;
root = insert(root, 50);
root = insert(root, 30);
root = insert(root, 20);
root = insert(root, 40);
root = insert(root, 70);
root = insert(root, 60);
root = insert(root, 80);
document.write("Before Reverse :" + "<br>");
// Print inorder traversal of the BST
inorder(root);
document.write("<br>");
// Reverse path till k
reversePath(root);
document.write("After Reverse :" + "<br>");
// Print inorder of reverse path tree
inorder(root);
// This code is contributed by itsok.
</script>
OutputBefore Reverse :
20 30 40 50 60 70 80
After Reverse :
20 30 40 80 60 70 50
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(n)
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