Swap Nodes [Algo]


Problem Statement :


A binary tree is a tree which is characterized by one of the following properties:

It can be empty (null).
It contains a root node only.
It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees.
In-order traversal is performed as

Traverse the left subtree.
Visit root.
Traverse the right subtree.
For this in-order traversal, start from the left child of the root node and keep exploring the left subtree until you reach a leaf. When you reach a leaf, back up to its parent, check for a right child and visit it if there is one. If there is not a child, you've explored its left and right subtrees fully. If there is a right child, traverse its left subtree then its right in the same manner. Keep doing this until you have traversed the entire tree. You will only store the values of a node as you visit when one of the following is true:

it is the first node visited, the first time visited
it is a leaf, should only be visited once
all of its subtrees have been explored, should only be visited once while this is true
it is the root of the tree, the first time visited
Swapping: Swapping subtrees of a node means that if initially node has left subtree L and right subtree R, then after swapping, the left subtree will be R and the right subtree, L.

For example, in the following tree, we swap children of node 1.

                                Depth
    1               1            [1]
   / \             / \
  2   3     ->    3   2          [2]
   \   \           \   \
    4   5           5   4        [3]
In-order traversal of left tree is 2 4 1 3 5 and of right tree is 3 5 1 2 4.

Swap operation:

We define depth of a node as follows:

The root node is at depth 1.
If the depth of the parent node is d, then the depth of current node will be d+1.
Given a tree and an integer, k, in one operation, we need to swap the subtrees of all the nodes at each depth h, where h ∈ [k, 2k, 3k,...]. In other words, if h is a multiple of k, swap the left and right subtrees of that level.

You are given a tree of n nodes where nodes are indexed from [1..n] and it is rooted at 1. You have to perform t swap operations on it, and after each swap operation print the in-order traversal of the current state of the tree.

Function Description

Complete the swapNodes function in the editor below. It should return a two-dimensional array where each element is an array of integers representing the node indices of an in-order traversal after a swap operation.

swapNodes has the following parameter(s):
- indexes: an array of integers representing index values of each node[i] , beginning with node[1], the first element, as the root.
- queries: an array of integers, each representing a k value.

Input Format
The first line contains n, number of nodes in the tree.

Each of the next n lines contains two integers, a b, where a is the index of left child, and b is the index of right child of ith node.

Note: -1 is used to represent a null node.

The next line contains an integer, t, the size of queries.
Each of the next t lines contains an integer queries[i], each being a value k.

Output Format
For each k, perform the swap operation and store the indices of your in-order traversal to your result array. After all swap operations have been performed, return your result array for printing.



Solution :


                            Solution in C :

In C ++ :



#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
vector<int> leftNode, rightNode;
int swapLevel;

void traverse(int node=1){
    if (node == -1) return;
    traverse(leftNode[node]);
    cout << node << " ";
    traverse(rightNode[node]);
    if (node == 1) cout << endl;
}

void swap(int level=1, int node=1) {
	if (node == -1) return;
	if (level % swapLevel == 0) {
		int tmp = leftNode[node];
		leftNode[node] = rightNode[node];
		rightNode[node] = tmp;
	}
	swap(level+1, leftNode[node]);
	swap(level+1, rightNode[node]);
}

int main() {
    int count;    
    cin>>count;
	leftNode.push_back(0);
    rightNode.push_back(0);
    while(count--){
        int L, R;
        cin>>L>>R;
        leftNode.push_back(L);
        rightNode.push_back(R);
    }
    cin>>count;
    while(count--){
		cin >> swapLevel;
		swap();
		traverse();
	}
    return 0;
}





In Java :




import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        Node root = buildTree(in);
        int t = in.nextInt();
        for(int i=0;i<t;++i){
            swapKDevisible(root, in.nextInt());
            inOrderTraversal(root);
            System.out.println();
        }
    }
    
    public static void swapKDevisible(Node nd, int k){
        if(nd==null)
            return;
        if(nd.depth%k==0){
            swapChildren(nd);
         }
        swapKDevisible(nd.left,k);
        swapKDevisible(nd.right,k);
    }
    
    public static Node buildTree(Scanner sc){
        HashMap<Integer, Node> cache = new HashMap<Integer, Node>();
        Node root = new Node(null, null, 1, 1);
        cache.put(1,root);
        int len = sc.nextInt();
        for(int i=0;i<len;++i){
            int ind1 = sc.nextInt();
            int ind2 = sc.nextInt();
            Node parent = cache.get(i+1);
            Node left = buildNewNode(ind1,parent.depth+1);
            Node right = buildNewNode(ind2,parent.depth+1);
            if(ind1!=-1)
                cache.put(ind1,left);
            if(ind2!=-1)
                cache.put(ind2,right);
            parent.left=left;
            parent.right=right;
        }
        return root;
    }
    
    public static Node buildNewNode(int data,int depth) {
        if(data==-1) return null;
        return new Node(null,null,data,depth);
    }
    
    public static void swapChildren(Node nd){
        if(nd==null)
            return;
        Node t = nd.right;
        nd.right = nd.left;
        nd.left = t;
    }
    
    public static void inOrderTraversal(Node nd){
        if(nd==null)
            return;
        inOrderTraversal(nd.left);
        System.out.print(nd.data+" ");
        inOrderTraversal(nd.right);
    }
    
    public static class Node{
        public Node left;
        public Node right;
        public int data;
        public int depth;
        
        public Node(Node left, Node right, int data, int depth){
            this.left=left;
            this.right=right;
            this.data=data;
            this.depth=depth;
        }
    }
}






In Pytho3 :



def inorder(T) :
    stack = [1]
    result = []
    while stack :
        i = stack.pop()
        if i > 0 :
            if T[i][1] > 0 : 
                stack.append(T[i][1])
            stack.append(-i)
            if T[i][0] > 0 : 
                stack.append(T[i][0])
        else :
            result.append(-i)
    return result

def swap(T, K) :
    toVisit, depth = [1], 1
    while toVisit :
        if depth % K == 0 :
            for i in toVisit :
                T[i] = (T[i][1], T[i][0])
        toVisit_ = []
        for i in toVisit :
            if T[i][0] > 0 :
                toVisit_.append(T[i][0])
            if T[i][1] > 0 :
                toVisit_.append(T[i][1])
        toVisit = toVisit_
        depth += 1

N = int(input())
T = [None] + [tuple(int(_) for _ in input().split()) for _ in range(N)]

N = int(input())
for _ in range(N) :
    swap(T,int(input()))
    print(" ".join(str(_) for _ in inorder(T)))
    




In C :



#include<stdio.h>
#include<stdlib.h>
#define S(a) scanf("%d",&a)
#define SS(a,b) scanf("%d%d",&a,&b)
struct node {
	int data;
	int left;
	int right;	
};

void swap(int *a, int *b){
	int temp = *a;
	*a = *b;
	*b = temp;
}
void swapNodes(struct node *arr,int i,int level,int k){
		if(i==-1){
			return;	
		}
		else {
			if(level==k){
				swap(&arr[i].left,&arr[i].right);	
				return;
			}
			else if (level<k){
				int left = arr[i].left;
				int right = arr[i].right;
				level++;	
				swapNodes(arr,left,level,k);
				swapNodes(arr,right,level,k);	
				return;	
			}
			else {
				return;	
			}
		}

}

int cDepth(struct node *arr,int i){
	if(i == -1){
			return 0;
	}
	int left = arr[i].left;
	int right = arr[i].right;
	int l = cDepth(arr,left);
	int r = cDepth(arr,right);
	if(l>r) { return l+1;}
	else 	{ return r+1;}
}

void inorderTr(struct node *arr,int i){
	if(i==-1){
		return;	
	}	
	else{
		int left = arr[i].left;
		int right = arr[i].right;
		inorderTr(arr,left);
		printf("%d ",i);
		inorderTr(arr,right);
	}
}

int main(){
	int i,n,t,k;
	S(n);
	struct node *arr = (struct node *)(malloc(sizeof(struct node) * (n+1)));
	for(i=1;i<=n;i++){
		arr[i].data = i;
		SS(arr[i].left,arr[i].right);
	}
	S(t);
	int j=1;
	int depth = cDepth(arr,1);
		
	while(t-->0) {
		S(k);
		j=1;
		//struct node *arrT = (struct node *)(malloc(sizeof(struct node) * (n+1)));	
		//copy original contents
		/*for(i=1;i<=n;i++){
			arrT[i].data = arr[i].data;
			arrT[i].left = arr[i].left;
			arrT[i].right = arr[i].right;
		}*/
		int tempk = k;
		while(tempk<=depth){
			swapNodes(arr,1,1,tempk);
			j++;
			tempk = k * j;
		}
        inorderTr(arr,1);
        if(t!=0){ printf("\n"); }
        //free(arrT);
	}
	
	free(arr);
	//for(i=1;i<=n;i++){
		//arr[i].data = i+1;
	//	printf("%d\t%d\t%d\n",arr[i].data,arr[i].left,arr[i].right);
	//}
	
	return 0;
}
                        




View More Similar Problems

Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

View Solution →

Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →

Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

View Solution →

QHEAP1

This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

View Solution →

Jesse and Cookies

Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t

View Solution →

Find the Running Median

The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.

View Solution →