Given a binary tree containing n nodes. The problem is to replace each node in the binary tree with the sum of its inorder predecessor and inorder successor.
Examples:
Input : 1 / \ 2 3 / \ / \ 4 5 6 7 Output : 11 / \ 9 13 / \ / \ 2 3 4 3 For 1: Inorder predecessor = 5 Inorder successor = 6 Sum = 11 For 4: Inorder predecessor = 0 (as inorder predecessor is not present) Inorder successor = 2 Sum = 2 For 7: Inorder predecessor = 3 Inorder successor = 0 (as inorder successor is not present) Sum = 3
Approach:
Create an array arr. Store 0 at index 0. Now, store the inorder traversal of tree in the array arr. Then, store 0 at last index. 0’s are stored as inorder predecessor of leftmost leaf and inorder successor of rightmost leaf is not present. Now, perform inorder traversal and while traversing node replace node’s value with arr[i-1] + arr[i+1] and then increment i.
In the beginning initialize i = 1. For an element arr[i], the values arr[i-1] and arr[i+1] are its inorder predecessor and inorder successor respectively.
Implementation:
// C++ implementation to replace each node // in binary tree with the sum of its inorder // predecessor and successor #include <bits/stdc++.h> using namespace std;
// node of a binary tree struct Node {
int data;
struct Node* left, *right;
}; // function to get a new node of a binary tree struct Node* getNode( int data)
{ // allocate node
struct Node* new_node =
( struct Node*) malloc ( sizeof ( struct Node));
// put in the data;
new_node->data = data;
new_node->left = new_node->right = NULL;
return new_node;
} // function to store the inorder traversal // of the binary tree in 'arr' void storeInorderTraversal( struct Node* root,
vector< int >& arr)
{ // if root is NULL
if (!root)
return ;
// first recur on left child
storeInorderTraversal(root->left, arr);
// then store the root's data in 'arr'
arr.push_back(root->data);
// now recur on right child
storeInorderTraversal(root->right, arr);
} // function to replace each node with the sum of its // inorder predecessor and successor void replaceNodeWithSum( struct Node* root,
vector< int > arr, int * i)
{ // if root is NULL
if (!root)
return ;
// first recur on left child
replaceNodeWithSum(root->left, arr, i);
// replace node's data with the sum of its
// inorder predecessor and successor
root->data = arr[*i - 1] + arr[*i + 1];
// move 'i' to point to the next 'arr' element
++*i;
// now recur on right child
replaceNodeWithSum(root->right, arr, i);
} // Utility function to replace each node in binary // tree with the sum of its inorder predecessor // and successor void replaceNodeWithSumUtil( struct Node* root)
{ // if tree is empty
if (!root)
return ;
vector< int > arr;
// store the value of inorder predecessor
// for the leftmost leaf
arr.push_back(0);
// store the inorder traversal of the tree in 'arr'
storeInorderTraversal(root, arr);
// store the value of inorder successor
// for the rightmost leaf
arr.push_back(0);
// replace each node with the required sum
int i = 1;
replaceNodeWithSum(root, arr, &i);
} // function to print the preorder traversal // of a binary tree void preorderTraversal( struct Node* root)
{ // if root is NULL
if (!root)
return ;
// first print the data of node
cout << root->data << " " ;
// then recur on left subtree
preorderTraversal(root->left);
// now recur on right subtree
preorderTraversal(root->right);
} // Driver program to test above int main()
{ // binary tree formation
struct Node* root = getNode(1); /* 1 */
root->left = getNode(2); /* / \ */
root->right = getNode(3); /* 2 3 */
root->left->left = getNode(4); /* / \ / \ */
root->left->right = getNode(5); /* 4 5 6 7 */
root->right->left = getNode(6);
root->right->right = getNode(7);
cout << "Preorder Traversal before tree modification:n" ;
preorderTraversal(root);
replaceNodeWithSumUtil(root);
cout << "\nPreorder Traversal after tree modification:n" ;
preorderTraversal(root);
return 0;
} |
// Java implementation to replace each node // in binary tree with the sum of its inorder // predecessor and successor import java.util.*;
class Solution
{ // node of a binary tree static class Node {
int data;
Node left, right;
} //INT class static class INT
{ int data;
} // function to get a new node of a binary tree static Node getNode( int data)
{ // allocate node
Node new_node = new Node();
// put in the data;
new_node.data = data;
new_node.left = new_node.right = null ;
return new_node;
} // function to store the inorder traversal // of the binary tree in 'arr' static void storeInorderTraversal( Node root,
Vector<Integer> arr)
{ // if root is null
if (root== null )
return ;
// first recur on left child
storeInorderTraversal(root.left, arr);
// then store the root's data in 'arr'
arr.add(root.data);
// now recur on right child
storeInorderTraversal(root.right, arr);
} // function to replace each node with the sum of its // inorder predecessor and successor static void replaceNodeWithSum( Node root,
Vector<Integer> arr, INT i)
{ // if root is null
if (root== null )
return ;
// first recur on left child
replaceNodeWithSum(root.left, arr, i);
// replace node's data with the sum of its
// inorder predecessor and successor
root.data = arr.get(i.data - 1 ) + arr.get(i.data + 1 );
// move 'i' to point to the next 'arr' element
i.data++;
// now recur on right child
replaceNodeWithSum(root.right, arr, i);
} // Utility function to replace each node in binary // tree with the sum of its inorder predecessor // and successor static void replaceNodeWithSumUtil( Node root)
{ // if tree is empty
if (root== null )
return ;
Vector<Integer> arr= new Vector<Integer>();
// store the value of inorder predecessor
// for the leftmost leaf
arr.add( 0 );
// store the inorder traversal of the tree in 'arr'
storeInorderTraversal(root, arr);
// store the value of inorder successor
// for the rightmost leaf
arr.add( 0 );
// replace each node with the required sum
INT i = new INT();
i.data= 1 ;
replaceNodeWithSum(root, arr, i);
} // function to print the preorder traversal // of a binary tree static void preorderTraversal( Node root)
{ // if root is null
if (root== null )
return ;
// first print the data of node
System.out.print( root.data + " " );
// then recur on left subtree
preorderTraversal(root.left);
// now recur on right subtree
preorderTraversal(root.right);
} // Driver program to test above public static void main(String args[])
{ // binary tree formation
Node root = getNode( 1 ); // 1
root.left = getNode( 2 ); // / \
root.right = getNode( 3 ); // 2 3
root.left.left = getNode( 4 ); // / \ / \
root.left.right = getNode( 5 ); // 4 5 6 7
root.right.left = getNode( 6 );
root.right.right = getNode( 7 );
System.out.println( "Preorder Traversal before tree modification:" );
preorderTraversal(root);
replaceNodeWithSumUtil(root);
System.out.println( "\nPreorder Traversal after tree modification:" );
preorderTraversal(root);
} } //contributed by Arnab Kundu |
# Python3 implementation to replace each # node in binary tree with the sum of its # inorder predecessor and successor # class to get a new node of a # binary tree class getNode:
def __init__( self , data):
# put in the data
self .data = data
self .left = self .right = None
# function to store the inorder traversal # of the binary tree in 'arr' def storeInorderTraversal(root, arr):
# if root is None
if ( not root):
return
# first recur on left child
storeInorderTraversal(root.left, arr)
# then store the root's data in 'arr'
arr.append(root.data)
# now recur on right child
storeInorderTraversal(root.right, arr)
# function to replace each node with the # sum of its inorder predecessor and successor def replaceNodeWithSum(root, arr, i):
# if root is None
if ( not root):
return
# first recur on left child
replaceNodeWithSum(root.left, arr, i)
# replace node's data with the sum of its
# inorder predecessor and successor
root.data = arr[i[ 0 ] - 1 ] + arr[i[ 0 ] + 1 ]
# move 'i' to point to the next 'arr' element
i[ 0 ] + = 1
# now recur on right child
replaceNodeWithSum(root.right, arr, i)
# Utility function to replace each node in # binary tree with the sum of its inorder # predecessor and successor def replaceNodeWithSumUtil(root):
# if tree is empty
if ( not root):
return
arr = []
# store the value of inorder predecessor
# for the leftmost leaf
arr.append( 0 )
# store the inorder traversal of the
# tree in 'arr'
storeInorderTraversal(root, arr)
# store the value of inorder successor
# for the rightmost leaf
arr.append( 0 )
# replace each node with the required sum
i = [ 1 ]
replaceNodeWithSum(root, arr, i)
# function to print the preorder traversal # of a binary tree def preorderTraversal(root):
# if root is None
if ( not root):
return
# first print the data of node
print (root.data, end = " " )
# then recur on left subtree
preorderTraversal(root.left)
# now recur on right subtree
preorderTraversal(root.right)
# Driver Code if __name__ = = '__main__' :
# binary tree formation
root = getNode( 1 ) # 1
root.left = getNode( 2 ) # / \
root.right = getNode( 3 ) # 2 3
root.left.left = getNode( 4 ) # / \ / \
root.left.right = getNode( 5 ) # 4 5 6 7
root.right.left = getNode( 6 )
root.right.right = getNode( 7 )
print ( "Preorder Traversal before" ,
"tree modification:" )
preorderTraversal(root)
replaceNodeWithSumUtil(root)
print ()
print ( "Preorder Traversal after" ,
"tree modification:" )
preorderTraversal(root)
# This code is contributed by PranchalK |
// C# implementation to replace each // node in binary tree with the sum // of its inorder predecessor and successor using System;
using System.Collections.Generic;
class GFG
{ // node of a binary tree public class Node
{ public int data;
public Node left, right;
} // INT class public class INT
{ public int data;
} // function to get a new node // of a binary tree public static Node getNode( int data)
{ // allocate node
Node new_node = new Node();
// put in the data;
new_node.data = data;
new_node.left = new_node.right = null ;
return new_node;
} // function to store the inorder traversal // of the binary tree in 'arr' public static void storeInorderTraversal(Node root,
List< int > arr)
{ // if root is null
if (root == null )
{
return ;
}
// first recur on left child
storeInorderTraversal(root.left, arr);
// then store the root's data in 'arr'
arr.Add(root.data);
// now recur on right child
storeInorderTraversal(root.right, arr);
} // function to replace each node with // the sum of its inorder predecessor // and successor public static void replaceNodeWithSum(Node root,
List< int > arr, INT i)
{ // if root is null
if (root == null )
{
return ;
}
// first recur on left child
replaceNodeWithSum(root.left, arr, i);
// replace node's data with the
// sum of its inorder predecessor
// and successor
root.data = arr[i.data - 1] + arr[i.data + 1];
// move 'i' to point to the
// next 'arr' element
i.data++;
// now recur on right child
replaceNodeWithSum(root.right, arr, i);
} // Utility function to replace each // node in binary tree with the sum // of its inorder predecessor and successor public static void replaceNodeWithSumUtil(Node root)
{ // if tree is empty
if (root == null )
{
return ;
}
List< int > arr = new List< int >();
// store the value of inorder
// predecessor for the leftmost leaf
arr.Add(0);
// store the inorder traversal
// of the tree in 'arr'
storeInorderTraversal(root, arr);
// store the value of inorder successor
// for the rightmost leaf
arr.Add(0);
// replace each node with
// the required sum
INT i = new INT();
i.data = 1;
replaceNodeWithSum(root, arr, i);
} // function to print the preorder // traversal of a binary tree public static void preorderTraversal(Node root)
{ // if root is null
if (root == null )
{
return ;
}
// first print the data of node
Console.Write(root.data + " " );
// then recur on left subtree
preorderTraversal(root.left);
// now recur on right subtree
preorderTraversal(root.right);
} // Driver Code public static void Main( string [] args)
{ // binary tree formation
Node root = getNode(1); // 1
root.left = getNode(2); // / \
root.right = getNode(3); // 2 3
root.left.left = getNode(4); // / \ / \
root.left.right = getNode(5); // 4 5 6 7
root.right.left = getNode(6);
root.right.right = getNode(7);
Console.WriteLine( "Preorder Traversal " +
"before tree modification:" );
preorderTraversal(root);
replaceNodeWithSumUtil(root);
Console.WriteLine( "\nPreorder Traversal after " +
"tree modification:" );
preorderTraversal(root);
} } // This code is contributed by Shrikant13 |
<script> // Javascript implementation to replace each node // in binary tree with the sum of its inorder // predecessor and successor class Node { constructor(data)
{
this .left = null ;
this .right = null ;
this .data = data;
}
} // Function to get a new node of a // binary tree function getNode(data)
{ // Allocate node
let new_node = new Node(data);
return new_node;
} // Function to store the inorder traversal // of the binary tree in 'arr' function storeInorderTraversal(root, arr)
{ // If root is null
if (root == null )
return ;
// First recur on left child
storeInorderTraversal(root.left, arr);
// then store the root's data in 'arr'
arr.push(root.data);
// Now recur on right child
storeInorderTraversal(root.right, arr);
} // Function to replace each node with the // sum of its inorder predecessor and successor function replaceNodeWithSum(root, arr)
{ // If root is null
if (root == null )
return ;
// First recur on left child
replaceNodeWithSum(root.left, arr);
// Replace node's data with the sum of its
// inorder predecessor and successor
root.data = arr[data - 1] + arr[data + 1];
// Move 'i' to point to the next 'arr' element
data++;
// Now recur on right child
replaceNodeWithSum(root.right, arr);
} // Utility function to replace each node in binary // tree with the sum of its inorder predecessor // and successor function replaceNodeWithSumUtil(root)
{ // If tree is empty
if (root == null )
return ;
let arr = [];
// Store the value of inorder predecessor
// for the leftmost leaf
arr.push(0);
// Store the inorder traversal of
// the tree in 'arr'
storeInorderTraversal(root, arr);
// Store the value of inorder successor
// for the rightmost leaf
arr.push(0);
// Replace each node with the required sum
data = 1;
replaceNodeWithSum(root, arr);
} // Function to print the preorder traversal // of a binary tree function preorderTraversal(root)
{ // If root is null
if (root == null )
return ;
// First print the data of node
document.write(root.data + " " );
// Then recur on left subtree
preorderTraversal(root.left);
// Now recur on right subtree
preorderTraversal(root.right);
} // Driver code // Binary tree formation let root = getNode(1); // 1
root.left = getNode(2); // / \
root.right = getNode(3); // 2 3
root.left.left = getNode(4); // / \ / \
root.left.right = getNode(5); // 4 5 6 7
root.right.left = getNode(6); root.right.right = getNode(7); document.write( "Preorder Traversal before " +
"tree modification:" + "</br>" );
preorderTraversal(root); replaceNodeWithSumUtil(root); document.write( "</br>" + "Preorder Traversal after " +
"tree modification:" + "</br>" );
preorderTraversal(root); // This code is contributed by divyeshrabadiya07 </script> |
Preorder Traversal before tree modification:n1 2 4 5 3 6 7 Preorder Traversal after tree modification:n11 9 2 3 13 4 3
Time Complexity: O(n), as we are traversing the tree having n nodes using recursion (preorder and inorder traversal).
Auxiliary Space: O(n), as we are using extra space for storing the inorder traversal.
Another Approach (without extra vector):
- Use inorder traversal and keep previous node and value of previous node’s previous.
- Update previous node with current node + previous value
Implementation:
// C++ implementation to replace each node // in binary tree with the sum of its inorder // predecessor and successor #include <bits/stdc++.h> using namespace std;
// node of a binary tree struct Node {
int data;
struct Node* left, *right;
}; // function to get a new node of a binary tree struct Node* getNode( int data)
{ // allocate node
struct Node* new_node =
( struct Node*) malloc ( sizeof ( struct Node));
// put in the data;
new_node->data = data;
new_node->left = new_node->right = NULL;
return new_node;
} // function to print the preorder traversal // of a binary tree void preorderTraversal( struct Node* root)
{ // if root is NULL
if (!root)
return ;
// first print the data of node
cout << root->data << " " ;
// then recur on left subtree
preorderTraversal(root->left);
// now recur on right subtree
preorderTraversal(root->right);
} void inOrderTraverse( struct Node* root, struct Node* &prev, int &prevVal)
{
if (root == NULL) return ;
inOrderTraverse(root->left, prev, prevVal);
if (prev == NULL)
{
prev = root;
prevVal = 0;
}
else
{
int temp = prev->data;
prev->data = prevVal + root->data;
prev = root;
prevVal = temp;
}
inOrderTraverse(root->right, prev, prevVal);
}
// Driver program to test above int main()
{ // binary tree formation
struct Node* root = getNode(1); /* 1 */
root->left = getNode(2); /* / \ */
root->right = getNode(3); /* 2 3 */
root->left->left = getNode(4); /* / \ / \ */
root->left->right = getNode(5); /* 4 5 6 7 */
root->right->left = getNode(6);
root->right->right = getNode(7);
cout << "Preorder Traversal before tree modification:\n" ;
preorderTraversal(root);
struct Node* prev = NULL;
int prevVal = -1;
inOrderTraverse(root, prev, prevVal);
// update rightmost node.
prev->data = prevVal;
cout << "\nPreorder Traversal after tree modification:\n" ;
preorderTraversal(root);
return 0;
} |
// Java implementation to replace each node // in binary tree with the sum of its inorder // predecessor and successor import java.util.*;
class Main {
// Driver program to test above
public static void main(String[] args)
{
GFG ob = new GFG();
// binary tree formation
Node root = ob.getNode( 1 ); // 1
root.left = ob.getNode( 2 ); // / \
root.right = ob.getNode( 3 ); // 2 3
root.left.left = ob.getNode( 4 ); // / \ / \
root.left.right = ob.getNode( 5 ); //4 5 6 7
root.right.left = ob.getNode( 6 );
root.right.right = ob.getNode( 7 );
System.out.println( "Preorder Traversal "
+ "before tree modification:" );
ob.preorderTraversal(root);
ob.inOrderTraverse(root);
// update rightmost node.
ob.prev.data = ob.prevVal;
System.out.println( "\nPreorder Traversal after "
+ "tree modification:" );
ob.preorderTraversal(root);
}
} // node of a binary tree class Node {
int data;
Node left, right;
} class GFG {
public static Node prev;
public static int prevVal;
// function to get a new node
// of a binary tree
static Node getNode( int data)
{
// allocate node
Node new_node = new Node();
// put in the data;
new_node.data = data;
new_node.left = new_node.right = null ;
return new_node;
}
// function to print the preorder
// traversal of a binary tree
static void preorderTraversal(Node root)
{
// if root is null
if (root == null ) {
return ;
}
// first print the data of node
System.out.print(root.data + " " );
// then recur on left subtree
preorderTraversal(root.left);
// now recur on right subtree
preorderTraversal(root.right);
}
static void inOrderTraverse(Node root)
{
if (root == null )
return ;
inOrderTraverse(root.left);
if (prev == null ) {
prev = root;
prevVal = 0 ;
}
else {
int temp = prev.data;
prev.data = prevVal + root.data;
prev = root;
prevVal = temp;
}
inOrderTraverse(root.right);
}
} // This code is contributed by Tapesh(tapeshdua420) |
# Python implementation to replace each node # in binary tree with the sum of its inorder # predecessor and successor # node of a binary tree class Node:
def __init__( self , data):
self .data = data
self .left = None
self .right = None
prev = None
prevVal = 0
# function to get a new node # of a binary tree def getNode(data):
# allocate node
new_node = Node(data)
return new_node
# function to print the preorder # traversal of a binary tree def preorderTraversal(root):
# if root is null
if root is None :
return
# first print the data of node
print (root.data, end = " " )
# then recur on left subtree
preorderTraversal(root.left)
# now recur on right subtree
preorderTraversal(root.right)
def inOrderTraverse(root):
global prev, prevVal
if root is None :
return
inOrderTraverse(root.left)
if prev is None :
prev = root
prevVal = 0
else :
temp = prev.data
prev.data = prevVal + root.data
prev = root
prevVal = temp
inOrderTraverse(root.right)
# Driver program to test above if __name__ = = '__main__' :
# binary tree formation
root = getNode( 1 ) # 1
root.left = getNode( 2 ) # / \
root.right = getNode( 3 ) # 2 3
root.left.left = getNode( 4 ) # / \ / \
root.left.right = getNode( 5 ) # 4 5 6 7
root.right.left = getNode( 6 )
root.right.right = getNode( 7 )
print ( "Preorder Traversal before tree modification:" )
preorderTraversal(root)
inOrderTraverse(root)
# update rightmost node.
prev.data = prevVal
print ( "\nPreorder Traversal after tree modification:" )
preorderTraversal(root)
# This code is contributed by Tapesh(tapeshdua420) |
// C# implementation to replace each node // in binary tree with the sum of its inorder // predecessor and successor using System;
using System.Collections.Generic;
class GFG {
// node of a binary tree
public class Node {
public int data;
public Node left, right;
}
// function to get a new node
// of a binary tree
public static Node getNode( int data)
{
// allocate node
Node new_node = new Node();
// put in the data;
new_node.data = data;
new_node.left = new_node.right = null ;
return new_node;
}
// function to print the preorder
// traversal of a binary tree
public static void preorderTraversal(Node root)
{
// if root is null
if (root == null ) {
return ;
}
// first print the data of node
Console.Write(root.data + " " );
// then recur on left subtree
preorderTraversal(root.left);
// now recur on right subtree
preorderTraversal(root.right);
}
public static void inOrderTraverse(Node root,
ref Node prev,
ref int prevVal)
{
if (root == null )
return ;
inOrderTraverse(root.left, ref prev, ref prevVal);
if (prev == null ) {
prev = root;
prevVal = 0;
}
else {
int temp = prev.data;
prev.data = prevVal + root.data;
prev = root;
prevVal = temp;
}
inOrderTraverse(root.right, ref prev, ref prevVal);
}
// Driver Code
public static void Main( string [] args)
{
// binary tree formation
Node root = getNode(1); // 1
root.left = getNode(2); // / \
root.right = getNode(3); // 2 3
root.left.left = getNode(4); // / \ / \
root.left.right = getNode(5); // 4 5 6 7
root.right.left = getNode(6);
root.right.right = getNode(7);
Console.WriteLine( "Preorder Traversal "
+ "before tree modification:" );
preorderTraversal(root);
Node prev = null ;
int prevVal = -1;
inOrderTraverse(root, ref prev, ref prevVal);
// update rightmost node.
prev.data = prevVal;
Console.WriteLine( "\nPreorder Traversal after "
+ "tree modification:" );
preorderTraversal(root);
}
} // This code is contributed by Abhijeet Kumar(abhijeet19403) |
// Python implementation to replace each node // in binary tree with the sum of its inorder // predecessor and successor // node of a binary tree class Node { constructor(data) {
this .data = data;
this .left = null ;
this .right = null ;
}
} let prev = null ;
let prevVal = 0; // function to get a new node // of a binary tree function getNode(data) {
// allocate node
let new_node = new Node(data);
return new_node;
} // function to print the preorder // traversal of a binary tree function preorderTraversal(root) {
// if root is null
if (root == null ) {
return ;
}
// first print the data of node
process.stdout.write(root.data + " " );
// then recur on left subtree
preorderTraversal(root.left);
// now recur on right subtree
preorderTraversal(root.right);
} function inOrderTraverse(root) {
if (root == null ) {
return ;
}
inOrderTraverse(root.left);
if (prev == null ) {
prev = root;
prevVal = 0;
} else {
let temp = prev.data;
prev.data = prevVal + root.data;
prev = root;
prevVal = temp;
}
inOrderTraverse(root.right);
} // Driver program to test above // binary tree formation let root = getNode(1); // 1
root.left = getNode(2); // / \
root.right = getNode(3); // 2 3
root.left.left = getNode(4); // / \ / \
root.left.right = getNode(5); // 4 5 6 7
root.right.left = getNode(6); root.right.right = getNode(7); console.log( "Preorder Traversal before tree modification:" );
preorderTraversal(root); inOrderTraverse(root); // update rightmost node. prev.data = prevVal; console.log( "\nPreorder Traversal after tree modification:" );
preorderTraversal(root); // This code is contributed by Tapesh(tapeshdua420) |
Preorder Traversal before tree modification: 1 2 4 5 3 6 7 Preorder Traversal after tree modification: 11 9 2 3 13 4 3
Time Complexity: O(n), as we are traversing the tree having n nodes using recursion.
Auxiliary Space: O(n), we are not using any explicit space but as we are using recursion there will be extra space due to recursive stack space.