Given a binary tree, the task is to find the distance between two keys in a binary tree, no parent pointers are given. The distance between two nodes is the minimum number of edges to be traversed to reach one node from other.
We have already discussed a method which uses segment tree to reduce the query time to O(logn), here the task is to reduce query time to O(1) by compromising space complexity to O(n logn). In this post, we will use Sparse table instead of segment tree for finding the minimum in given range, which uses dynamic programming and bit manipulation to achieve O(1) query time.
A sparse table will preprocess the minimum values of given range for L array in Nlogn space i.e. each node will contain chain of values of log(i) length where i is the index of the ith node in L array. Each entry in the sparse table says M[i][j] will represent the index of the minimum value in the subarray starting at i having length 2^j.
The distance between two nodes can be obtained in terms of lowest common ancestor.
Dist(n1, n2) = Level[n1] + Level[n2] - 2*Level[lca]
This problem can be breakdown into:
- Finding levels of each node
- Finding the Euler tour of binary tree
- Building sparse table for LCA.
These steps are explained below :
- Find the levels of each node by applying level order traversal.
- Find the LCA of two nodes in binary tree in O(logn) by Storing Euler tour of Binary tree in array and computing two other arrays with the help of levels of each node and Euler tour.
These steps are shown below:
- First, find Euler Tour of binary tree.
- Then, store levels of each node in Euler array.
- Then, store First occurrences of all nodes of binary tree in Euler array. H stores the indices of nodes from Euler array, so that range of query for finding minimum can be minimized and their by further optimizing the query time.
- Then build sparse table on L array and find the minimum value say X in range (H[A] to H[B]). Then, we use the index of value X as an index to Euler array to get LCA, i.e. Euler[index(X)].
Let, A=8 and B=5.- H[8]= 1 and H[5]=2
- we get min value in L array between 1 and 2 as X=0, index=7
- Then, LCA= Euler[7], i.e LCA=1.
- Finally, apply distance formula discussed above to get the distance between two nodes.
Implementation:
#include <bits/stdc++.h> #define MAX 100001 using namespace std;
/* A tree node structure */ struct Node {
int data;
struct Node* left;
struct Node* right;
}; /* Utility function to create a new Binary Tree node */ struct Node* newNode( int data)
{ struct Node* temp = new struct Node;
temp->data = data;
temp->left = temp->right = NULL;
return temp;
} // Array to store level of each node int level[MAX];
// Utility Function to store level of all nodes void FindLevels( struct Node* root)
{ if (!root)
return ;
// queue to hold tree node with level
queue<pair< struct Node*, int > > q;
// let root node be at level 0
q.push({ root, 0 });
pair< struct Node*, int > p;
// Do level Order Traversal of tree
while (!q.empty()) {
p = q.front();
q.pop();
// Node p.first is on level p.second
level[p.first->data] = p.second;
// If left child exits, put it in queue
// with current_level +1
if (p.first->left)
q.push({ p.first->left, p.second + 1 });
// If right child exists, put it in queue
// with current_level +1
if (p.first->right)
q.push({ p.first->right, p.second + 1 });
}
} // Stores Euler Tour int Euler[MAX];
// index in Euler array int idx = 0;
// Find Euler Tour void eulerTree( struct Node* root)
{ // store current node's data
Euler[++idx] = root->data;
// If left node exists
if (root->left) {
// traverse left subtree
eulerTree(root->left);
// store parent node's data
Euler[++idx] = root->data;
}
// If right node exists
if (root->right) {
// traverse right subtree
eulerTree(root->right);
// store parent node's data
Euler[++idx] = root->data;
}
} // checks for visited nodes int vis[MAX];
// Stores level of Euler Tour int L[MAX];
// Stores indices of the first occurrence // of nodes in Euler tour int H[MAX];
// Preprocessing Euler Tour for finding LCA void preprocessEuler( int size)
{ for ( int i = 1; i <= size; i++) {
L[i] = level[Euler[i]];
// If node is not visited before
if (vis[Euler[i]] == 0) {
// Add to first occurrence
H[Euler[i]] = i;
// Mark it visited
vis[Euler[i]] = 1;
}
}
} // Sparse table of size [MAX][LOGMAX] // M[i][j] is the index of the minimum value in // the sub array starting at i having length 2^j int M[MAX][18];
// Utility function to preprocess Sparse table void preprocessLCA( int N)
{ for ( int i = 0; i < N; i++)
M[i][0] = i;
for ( int j = 1; 1 << j <= N; j++)
for ( int i = 0; i + (1 << j) - 1 < N; i++)
if (L[M[i][j - 1]] < L[M[i + (1 << (j - 1))][j - 1]])
M[i][j] = M[i][j - 1];
else
M[i][j] = M[i + (1 << (j - 1))][j - 1];
} // Utility function to find the index of the minimum // value in range a to b int LCA( int a, int b)
{ // Subarray of length 2^j
int j = log2(b - a + 1);
if (L[M[a][j]] <= L[M[b - (1 << j) + 1][j]])
return M[a][j];
else
return M[b - (1 << j) + 1][j];
} // Function to return distance between // two nodes n1 and n2 int findDistance( int n1, int n2)
{ // Maintain original Values
int prevn1 = n1, prevn2 = n2;
// Get First Occurrence of n1
n1 = H[n1];
// Get First Occurrence of n2
n2 = H[n2];
// Swap if low>high
if (n2 < n1)
swap(n1, n2);
// Get position of minimum value
int lca = LCA(n1, n2);
// Extract value out of Euler tour
lca = Euler[lca];
// return calculated distance
return level[prevn1] + level[prevn2] - 2 * level[lca];
} void preProcessing(Node* root, int N)
{ // Build Tree
eulerTree(root);
// Store Levels
FindLevels(root);
// Find L and H array
preprocessEuler(2 * N - 1);
// Build sparse table
preprocessLCA(2 * N - 1);
} /* Driver function to test above functions */ int main()
{ // Number of nodes
int N = 8;
/* Constructing tree given in the above figure */
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->right->left->right = newNode(8);
// Function to do all preprocessing
preProcessing(root, N);
cout << "Dist(4, 5) = " << findDistance(4, 5) << "\n" ;
cout << "Dist(4, 6) = " << findDistance(4, 6) << "\n" ;
cout << "Dist(3, 4) = " << findDistance(3, 4) << "\n" ;
cout << "Dist(2, 4) = " << findDistance(2, 4) << "\n" ;
cout << "Dist(8, 5) = " << findDistance(8, 5) << "\n" ;
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG {
static class Pair<T, V> {
T first;
V second;
Pair() {
}
Pair(T first, V second) {
this .first = first;
this .second = second;
}
}
static class Node {
int data;
Node left, right;
Node( int data) {
this .data = data;
this .left = this .right = null ;
}
}
static int MAX = 100001 ;
// Array to store level of each node
static int [] level = new int [MAX];
// Utility Function to store level of all nodes
static void FindLevels(Node root) {
if (root == null )
return ;
// queue to hold tree node with level
Queue<Pair<Node, Integer>> q = new LinkedList<>();
// let root node be at level 0
q.add( new Pair<>(root, 0 ));
Pair<Node, Integer> p = new Pair<>();
// Do level Order Traversal of tree
while (!q.isEmpty()) {
p = q.poll();
// Node p.first is on level p.second
level[p.first.data] = p.second;
// If left child exits, put it in queue
// with current_level +1
if (p.first.left != null )
q.add( new Pair<>(p.first.left, p.second + 1 ));
// If right child exists, put it in queue
// with current_level +1
if (p.first.right != null )
q.add( new Pair<>(p.first.right, p.second + 1 ));
}
}
// Stores Euler Tour
static int [] Euler = new int [MAX];
// index in Euler array
static int idx = 0 ;
// Find Euler Tour
static void eulerTree(Node root) {
// store current node's data
Euler[++idx] = root.data;
// If left node exists
if (root.left != null ) {
// traverse left subtree
eulerTree(root.left);
// store parent node's data
Euler[++idx] = root.data;
}
// If right node exists
if (root.right != null ) {
// traverse right subtree
eulerTree(root.right);
// store parent node's data
Euler[++idx] = root.data;
}
}
// checks for visited nodes
static int [] vis = new int [MAX];
// Stores level of Euler Tour
static int [] L = new int [MAX];
// Stores indices of the first occurrence
// of nodes in Euler tour
static int [] H = new int [MAX];
// Preprocessing Euler Tour for finding LCA
static void preprocessEuler( int size) {
for ( int i = 1 ; i <= size; i++) {
L[i] = level[Euler[i]];
// If node is not visited before
if (vis[Euler[i]] == 0 ) {
// Add to first occurrence
H[Euler[i]] = i;
// Mark it visited
vis[Euler[i]] = 1 ;
}
}
}
// Sparse table of size [MAX][LOGMAX]
// M[i][j] is the index of the minimum value in
// the sub array starting at i having length 2^j
static int [][] M = new int [MAX][ 18 ];
// Utility function to preprocess Sparse table
static void preprocessLCA( int N) {
for ( int i = 0 ; i < N; i++)
M[i][ 0 ] = i;
for ( int j = 1 ; 1 << j <= N; j++)
for ( int i = 0 ; i + ( 1 << j) - 1 < N; i++)
if (L[M[i][j - 1 ]] < L[M[i + ( 1 << (j - 1 ))][j - 1 ]])
M[i][j] = M[i][j - 1 ];
else
M[i][j] = M[i + ( 1 << (j - 1 ))][j - 1 ];
}
// Utility function to find the index of the minimum
// value in range a to b
static int LCA( int a, int b) {
// Subarray of length 2^j
int j = ( int ) (Math.log(b - a + 1 ) / Math.log( 2 ));
if (L[M[a][j]] <= L[M[b - ( 1 << j) + 1 ][j]])
return M[a][j];
else
return M[b - ( 1 << j) + 1 ][j];
}
// Function to return distance between
// two nodes n1 and n2
static int findDistance( int n1, int n2) {
// Maintain original Values
int prevn1 = n1, prevn2 = n2;
// Get First Occurrence of n1
n1 = H[n1];
// Get First Occurrence of n2
n2 = H[n2];
// Swap if low>high
if (n2 < n1) {
int temp = n1;
n1 = n2;
n2 = temp;
}
// Get position of minimum value
int lca = LCA(n1, n2);
// Extract value out of Euler tour
lca = Euler[lca];
// return calculated distance
return level[prevn1] + level[prevn2] - 2 * level[lca];
}
static void preProcessing(Node root, int N) {
// Build Tree
eulerTree(root);
// Store Levels
FindLevels(root);
// Find L and H array
preprocessEuler( 2 * N - 1 );
// Build sparse table
preprocessLCA( 2 * N - 1 );
}
// Driver Code
public static void main(String[] args) {
// Number of nodes
int N = 8 ;
/* Constructing tree given in the above figure */
Node root = new Node( 1 );
root.left = new Node( 2 );
root.right = new Node( 3 );
root.left.left = new Node( 4 );
root.left.right = new Node( 5 );
root.right.left = new Node( 6 );
root.right.right = new Node( 7 );
root.right.left.right = new Node( 8 );
// Function to do all preprocessing
preProcessing(root, N);
System.out.println( "Dist(4, 5) = " + findDistance( 4 , 5 ));
System.out.println( "Dist(4, 6) = " + findDistance( 4 , 6 ));
System.out.println( "Dist(3, 4) = " + findDistance( 3 , 4 ));
System.out.println( "Dist(2, 4) = " + findDistance( 2 , 4 ));
System.out.println( "Dist(8, 5) = " + findDistance( 8 , 5 ));
}
} // This code is contributed by // sanjeev2552 |
from collections import deque
from math import log2
MAX = 100001
# A tree node structure class Node:
def __init__( self , data):
self .data = data
self .left = None
self .right = None
# Array to store level of each node level = [ 0 ] * MAX
# Utility Function to store level of all nodes def findLevels(root: Node):
global level
if root is None :
return
# queue to hold tree node with level
q = deque()
# let root node be at level 0
q.append((root, 0 ))
# Do level Order Traversal of tree
while q:
p = q[ 0 ]
q.popleft()
# Node p.first is on level p.second
level[p[ 0 ].data] = p[ 1 ]
# If left child exits, put it in queue
# with current_level +1
if p[ 0 ].left:
q.append((p[ 0 ].left, p[ 1 ] + 1 ))
# If right child exists, put it in queue
# with current_level +1
if p[ 0 ].right:
q.append((p[ 0 ].right, p[ 1 ] + 1 ))
# Stores Euler Tour Euler = [ 0 ] * MAX
# index in Euler array idx = 0
# Find Euler Tour def eulerTree(root: Node):
global Euler, idx
idx + = 1
# store current node's data
Euler[idx] = root.data
# If left node exists
if root.left:
# traverse left subtree
eulerTree(root.left)
idx + = 1
# store parent node's data
Euler[idx] = root.data
# If right node exists
if root.right:
# traverse right subtree
eulerTree(root.right)
idx + = 1
# store parent node's data
Euler[idx] = root.data
# checks for visited nodes vis = [ 0 ] * MAX
# Stores level of Euler Tour L = [ 0 ] * MAX
# Stores indices of the first occurrence # of nodes in Euler tour H = [ 0 ] * MAX
# Preprocessing Euler Tour for finding LCA def preprocessEuler(size: int ):
global L, H, vis
for i in range ( 1 , size + 1 ):
L[i] = level[Euler[i]]
# If node is not visited before
if vis[Euler[i]] = = 0 :
# Add to first occurrence
H[Euler[i]] = i
# Mark it visited
vis[Euler[i]] = 1
# Sparse table of size [MAX][LOGMAX] # M[i][j] is the index of the minimum value in # the sub array starting at i having length 2^j M = [[ 0 for i in range ( 18 )] for j in range ( MAX )]
# Utility function to preprocess Sparse table def preprocessLCA(N: int ):
global M
for i in range (N):
M[i][ 0 ] = i
j = 1
while 1 << j < = N:
i = 0
while i + ( 1 << j) - 1 < N:
if L[M[i][j - 1 ]] < L[M[i +
( 1 << (j - 1 ))][j - 1 ]]:
M[i][j] = M[i][j - 1 ]
else :
M[i][j] = M[i + ( 1 << (j - 1 ))][j - 1 ]
i + = 1
j + = 1
# Utility function to find the index of the minimum # value in range a to b def LCA(a: int , b: int ) - > int :
# Subarray of length 2^j
j = int (log2(b - a + 1 ))
if L[M[a][j]] < = L[M[b - ( 1 << j) + 1 ][j]]:
return M[a][j]
else :
return M[b - ( 1 << j) + 1 ][j]
# Function to return distance between # two nodes n1 and n2 def findDistance(n1: int , n2: int ) - > int :
# Maintain original Values
prevn1 = n1
prevn2 = n2
# Get First Occurrence of n1
n1 = H[n1]
# Get First Occurrence of n2
n2 = H[n2]
# Swap if low>high
if n2 < n1:
n1, n2 = n2, n1
# Get position of minimum value
lca = LCA(n1, n2)
# Extract value out of Euler tour
lca = Euler[lca]
# return calculated distance
return level[prevn1] + level[prevn2] - 2 * level[lca]
def preProcessing(root: Node, N: int ):
# Build Tree
eulerTree(root)
# Store Levels
findLevels(root)
# Find L and H array
preprocessEuler( 2 * N - 1 )
# Build sparse table
preprocessLCA( 2 * N - 1 )
# Driver Code if __name__ = = "__main__" :
# Number of nodes
N = 8
# Constructing tree given in the above figure
root = Node( 1 )
root.left = Node( 2 )
root.right = Node( 3 )
root.left.left = Node( 4 )
root.left.right = Node( 5 )
root.right.left = Node( 6 )
root.right.right = Node( 7 )
root.right.left.right = Node( 8 )
# Function to do all preprocessing
preProcessing(root, N)
print ( "Dist(4, 5) =" , findDistance( 4 , 5 ))
print ( "Dist(4, 6) =" , findDistance( 4 , 6 ))
print ( "Dist(3, 4) =" , findDistance( 3 , 4 ))
print ( "Dist(2, 4) =" , findDistance( 2 , 4 ))
print ( "Dist(8, 5) =" , findDistance( 8 , 5 ))
# This code is contributed by # sanjeev2552 |
// C# implementation of the approach using System;
using System.Collections.Generic;
public class GFG
{ public
class Pair<T, V>
{
public
T first;
public
V second;
public
Pair() {
}
public
Pair(T first, V second)
{
this .first = first;
this .second = second;
}
}
public
class Node
{
public
int data;
public
Node left, right;
public
Node( int data)
{
this .data = data;
this .left = this .right = null ;
}
}
static int MAX = 100001;
// Array to store level of each node
static int [] level = new int [MAX];
// Utility Function to store level of all nodes
static void FindLevels(Node root)
{
if (root == null )
return ;
// queue to hold tree node with level
Queue<Pair<Node, int >> q = new Queue<Pair<Node, int >>();
// let root node be at level 0
q.Enqueue( new Pair<Node, int >(root, 0));
Pair<Node, int > p = new Pair<Node, int >();
// Do level Order Traversal of tree
while (q.Count != 0)
{
p = q.Peek();
q.Dequeue();
// Node p.first is on level p.second
level[p.first.data] = p.second;
// If left child exits, put it in queue
// with current_level +1
if (p.first.left != null )
q.Enqueue( new Pair<Node, int >(p.first.left, p.second + 1));
// If right child exists, put it in queue
// with current_level +1
if (p.first.right != null )
q.Enqueue( new Pair<Node, int >(p.first.right, p.second + 1));
}
}
// Stores Euler Tour
static int [] Euler = new int [MAX];
// index in Euler array
static int idx = 0;
// Find Euler Tour
static void eulerTree(Node root)
{
// store current node's data
Euler[++idx] = root.data;
// If left node exists
if (root.left != null )
{
// traverse left subtree
eulerTree(root.left);
// store parent node's data
Euler[++idx] = root.data;
}
// If right node exists
if (root.right != null )
{
// traverse right subtree
eulerTree(root.right);
// store parent node's data
Euler[++idx] = root.data;
}
}
// checks for visited nodes
static int [] vis = new int [MAX];
// Stores level of Euler Tour
static int [] L = new int [MAX];
// Stores indices of the first occurrence
// of nodes in Euler tour
static int [] H = new int [MAX];
// Preprocessing Euler Tour for finding LCA
static void preprocessEuler( int size) {
for ( int i = 1; i <= size; i++) {
L[i] = level[Euler[i]];
// If node is not visited before
if (vis[Euler[i]] == 0)
{
// Add to first occurrence
H[Euler[i]] = i;
// Mark it visited
vis[Euler[i]] = 1;
}
}
}
// Sparse table of size [MAX,LOGMAX]
// M[i,j] is the index of the minimum value in
// the sub array starting at i having length 2^j
static int [,] M = new int [MAX, 18];
// Utility function to preprocess Sparse table
static void preprocessLCA( int N)
{
for ( int i = 0; i < N; i++)
M[i, 0] = i;
for ( int j = 1; 1 << j <= N; j++)
for ( int i = 0; i + (1 << j) - 1 < N; i++)
if (L[M[i, j - 1]] < L[M[i + (1 << (j - 1)), j - 1]])
M[i, j] = M[i, j - 1];
else
M[i, j] = M[i + (1 << (j - 1)), j - 1];
}
// Utility function to find the index of the minimum
// value in range a to b
static int LCA( int a, int b)
{
// Subarray of length 2^j
int j = ( int ) (Math.Log(b - a + 1) / Math.Log(2));
if (L[M[a,j]] <= L[M[b - (1 << j) + 1,j]])
return M[a,j];
else
return M[b - (1 << j) + 1,j];
}
// Function to return distance between
// two nodes n1 and n2
static int findDistance( int n1, int n2) {
// Maintain original Values
int prevn1 = n1, prevn2 = n2;
// Get First Occurrence of n1
n1 = H[n1];
// Get First Occurrence of n2
n2 = H[n2];
// Swap if low>high
if (n2 < n1)
{
int temp = n1;
n1 = n2;
n2 = temp;
}
// Get position of minimum value
int lca = LCA(n1, n2);
// Extract value out of Euler tour
lca = Euler[lca];
// return calculated distance
return level[prevn1] + level[prevn2] - 2 * level[lca];
}
static void preProcessing(Node root, int N)
{
// Build Tree
eulerTree(root);
// Store Levels
FindLevels(root);
// Find L and H array
preprocessEuler(2 * N - 1);
// Build sparse table
preprocessLCA(2 * N - 1);
}
// Driver Code
public static void Main(String[] args)
{
// Number of nodes
int N = 8;
/* Constructing tree given in the above figure */
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.right.left.right = new Node(8);
// Function to do all preprocessing
preProcessing(root, N);
Console.WriteLine( "Dist(4, 5) = " + findDistance(4, 5));
Console.WriteLine( "Dist(4, 6) = " + findDistance(4, 6));
Console.WriteLine( "Dist(3, 4) = " + findDistance(3, 4));
Console.WriteLine( "Dist(2, 4) = " + findDistance(2, 4));
Console.WriteLine( "Dist(8, 5) = " + findDistance(8, 5));
}
} // This code is contributed by aashish1995 |
<script> // Javascript implementation of the approach class Pair{ constructor(first, second)
{
this .first = first;
this .second = second;
}
} class Node{ constructor(data)
{
this .data = data;
this .left = null ;
this .right = null ;
}
} var MAX = 100001;
// Array to store level of each node var level = Array(MAX);
// Utility Function to store level of all nodes function FindLevels(root)
{ if (root == null )
return ;
// queue to hold tree node with level
var q = [];
// let root node be at level 0
q.push( new Pair(root, 0));
var p = new Pair();
// Do level Order Traversal of tree
while (q.length != 0)
{
p = q[0];
q.shift();
// Node p.first is on level p.second
level[p.first.data] = p.second;
// If left child exits, put it in queue
// with current_level +1
if (p.first.left != null )
q.push( new Pair(p.first.left, p.second + 1));
// If right child exists, put it in queue
// with current_level +1
if (p.first.right != null )
q.push( new Pair(p.first.right, p.second + 1));
}
} // Stores Euler Tour var Euler = Array(MAX);
// index in Euler array var idx = 0;
// Find Euler Tour function eulerTree(root)
{ // store current node's data
Euler[++idx] = root.data;
// If left node exists
if (root.left != null )
{
// traverse left subtree
eulerTree(root.left);
// store parent node's data
Euler[++idx] = root.data;
}
// If right node exists
if (root.right != null )
{
// traverse right subtree
eulerTree(root.right);
// store parent node's data
Euler[++idx] = root.data;
}
} // checks for visited nodes var vis = Array(MAX).fill(0);
// Stores level of Euler Tour var L = Array(MAX).fill(0);
// Stores indices of the first occurrence // of nodes in Euler tour var H = Array(MAX).fill(0);
// Preprocessing Euler Tour for finding LCA function preprocessEuler(size)
{ for ( var i = 1; i <= size; i++)
{
L[i] = level[Euler[i]];
// If node is not visited before
if (vis[Euler[i]] == 0)
{
// Add to first occurrence
H[Euler[i]] = i;
// Mark it visited
vis[Euler[i]] = 1;
}
}
} // Sparse table of size [MAX,LOGMAX] // M[i,j] is the index of the minimum value in // the sub array starting at i having length 2^j var M = Array.from(Array(MAX), ()=>Array(18));
// Utility function to preprocess Sparse table function preprocessLCA(N)
{ for ( var i = 0; i < N; i++)
M[i][0] = i;
for ( var j = 1; 1 << j <= N; j++)
for ( var i = 0; i + (1 << j) - 1 < N; i++)
if (L[M[i][j - 1]] < L[M[i + (1 << (j - 1))][j - 1]])
M[i][j] = M[i][j - 1];
else
M[i][j] = M[i + (1 << (j - 1))][j - 1];
} // Utility function to find the index of the minimum // value in range a to b function LCA(a, b)
{ // Subarray of length 2^j
var j = parseInt(Math.log(b - a + 1) / Math.log(2));
if (L[M[a][j]] <= L[M[b - (1 << j) + 1][j]])
return M[a][j];
else
return M[b - (1 << j) + 1][j];
} // Function to return distance between // two nodes n1 and n2 function findDistance( n1, n2)
{ // Maintain original Values
var prevn1 = n1, prevn2 = n2;
// Get First Occurrence of n1
n1 = H[n1];
// Get First Occurrence of n2
n2 = H[n2];
// Swap if low>high
if (n2 < n1)
{
var temp = n1;
n1 = n2;
n2 = temp;
}
// Get position of minimum value
var lca = LCA(n1, n2);
// Extract value out of Euler tour
lca = Euler[lca];
// return calculated distance
return level[prevn1] + level[prevn2] - 2 * level[lca];
} function preProcessing(root, N)
{ // Build Tree
eulerTree(root);
// Store Levels
FindLevels(root);
// Find L and H array
preprocessEuler(2 * N - 1);
// Build sparse table
preprocessLCA(2 * N - 1);
} // Driver Code // Number of nodes var N = 8;
/* Constructing tree given in the above figure */ var root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.right.left.right = new Node(8);
// Function to do all preprocessing preProcessing(root, N); document.write( "Dist(4, 5) = " + findDistance(4, 5) + "<br>" );
document.write( "Dist(4, 6) = " + findDistance(4, 6) + "<br>" );
document.write( "Dist(3, 4) = " + findDistance(3, 4) + "<br>" );
document.write( "Dist(2, 4) = " + findDistance(2, 4) + "<br>" );
document.write( "Dist(8, 5) = " + findDistance(8, 5) + "<br>" );
// This code is contributed by itsok. </script> |
Dist(4, 5) = 2 Dist(4, 6) = 4 Dist(3, 4) = 3 Dist(2, 4) = 1 Dist(8, 5) = 5
Complexity Analysis:
- Time Complexity: O(N log N)
- Space Complexity: O(N log N)