Given a parent array P, where P[i] indicates the parent of the ith node in the tree(assume parent of root node id indicated with -1). Find the height of the tree.
Examples:
Input : array[] = [-1 0 1 6 6 0 0 2 7]
Output : height = 5
Tree formed is:
0
/ | \
5 1 6
/ | \
2 4 3
/
7
/
8
- Start at each node and keep going to its parent until we reach -1.
- Also, keep track of the maximum height between all nodes.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int findHeight(int* parent, int n)
{
int res = 0;
for (int i = 0; i < n; i++) {
int p = i, current = 1;
while (parent[p] != -1) {
current++;
p = parent[p];
}
res = max(res, current);
}
return res;
}
int main()
{
int parent[] = { -1, 0, 1, 6, 6, 0, 0, 2, 7 };
int n = sizeof(parent) / sizeof(parent[0]);
int height = findHeight(parent, n);
cout << "Height of the given tree is: "
<< height << endl;
return 0;
}
|
Java
import java.io.*;
public class GFG {
static int findHeight(int[] parent, int n)
{
int res = 0;
for (int i = 0; i < n; i++) {
int p = i, current = 1;
while (parent[p] != -1) {
current++;
p = parent[p];
}
res = Math.max(res, current);
}
return res;
}
static public void main(String[] args)
{
int[] parent = { -1, 0, 1, 6, 6, 0,
0, 2, 7 };
int n = parent.length;
int height = findHeight(parent, n);
System.out.println("Height of the "
+ "given tree is: " + height);
}
}
|
Python3
def findHeight(parent, n):
res = 0
for i in range(n):
p = i
current = 1
while (parent[p] != -1):
current+= 1
p = parent[p]
res = max(res, current)
return res
if __name__ == '__main__':
parent = [-1, 0, 1, 6, 6, 0, 0, 2, 7]
n = len(parent)
height = findHeight(parent, n)
print("Height of the given tree is:", height)
|
C#
using System;
public class GFG {
static int findHeight(int[] parent, int n)
{
int res = 0;
for (int i = 0; i < n; i++) {
int p = i, current = 1;
while (parent[p] != -1) {
current++;
p = parent[p];
}
res = Math.Max(res, current);
}
return res;
}
static public void Main()
{
int[] parent = { -1, 0, 1, 6, 6, 0,
0, 2, 7 };
int n = parent.Length;
int height = findHeight(parent, n);
Console.WriteLine("Height of the "
+ "given tree is: " + height);
}
}
|
Javascript
<script>
function findHeight(parent,n)
{
let res = 0;
for (let i = 0; i < n; i++) {
let p = i, current = 1;
while (parent[p] != -1) {
current++;
p = parent[p];
}
res = Math.max(res, current);
}
return res;
}
let parent=[-1, 0, 1, 6, 6, 0,
0, 2, 7];
let n = parent.length;
let height = findHeight(parent, n);
document.write("Height of the "
+ "given tree is: " + height);
</script>
|
Output:
Height of the given tree is: 5
Time Complexity : O( N^2 )
Space Complexity : O( 1 )
Optimized approach: We use dynamic programming. We store the height from root to each node in an array. So, if we know the height of the root to a node, then we can get the height from the root to the node child by simply adding 1.
Implementation:
CPP
#include <bits/stdc++.h>
using namespace std;
int rec(int i, int parent[], vector<int> height)
{
if (parent[i] == -1) {
return 1;
}
if (height[i] != -1) {
return height[i];
}
height[i] = rec(parent[i], parent, height) + 1;
return height[i];
}
int findHeight(int* parent, int n)
{
int res = 0;
vector<int> height(n, -1);
for (int i = 0; i < n; i++) {
res = max(res, rec(i, parent, height));
}
return res;
}
int main()
{
int parent[] = { -1, 0, 1, 6, 6, 0, 0, 2, 7 };
int n = sizeof(parent) / sizeof(parent[0]);
int height = findHeight(parent, n);
cout << "Height of the given tree is: "
<< height << endl;
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static int rec(int i, int parent[], int[] height)
{
if (parent[i] == -1) {
return 1;
}
if (height[i] != -1) {
return height[i];
}
height[i] = rec(parent[i], parent, height) + 1;
return height[i];
}
static int findHeight(int[] parent, int n)
{
int res = 0;
int height[]=new int[n];
Arrays.fill(height,-1);
for (int i = 0; i < n; i++) {
res = Math.max(res, rec(i, parent, height));
}
return res;
}
public static void main (String[] args) {
int[] parent = { -1, 0, 1, 6, 6, 0, 0, 2, 7 };
int n = parent.length;
int height = findHeight(parent, n);
System.out.println("Height of the given tree is: "+height);
}
}
|
Python3
def rec(i, parent, height):
if (parent[i] == -1):
return 1
if (height[i] != -1):
return height[i]
height[i] = rec(parent[i], parent, height) + 1
return height[i]
def findHeight(parent, n):
res = 0
height = [-1]*(n)
for i in range(n):
res = max(res, rec(i, parent, height))
return res
if __name__ == '__main__':
parent = [-1, 0, 1, 6, 6, 0, 0, 2, 7]
n = len(parent)
height = findHeight(parent, n)
print("Height of the given tree is: ",height)
|
C#
using System;
public class GFG{
static int rec(int i, int[] parent, int[] height)
{
if (parent[i] == -1) {
return 1;
}
if (height[i] != -1) {
return height[i];
}
height[i] = rec(parent[i], parent, height) + 1;
return height[i];
}
static int findHeight(int[] parent, int n)
{
int res = 0;
int[] height = new int[n];
Array.Fill(height, -1);
for (int i = 0; i < n; i++) {
res = Math.Max(res, rec(i, parent, height));
}
return res;
}
static public void Main ()
{
int[] parent = { -1, 0, 1, 6, 6, 0, 0, 2, 7 };
int n = parent.Length;
int height = findHeight(parent, n);
Console.WriteLine("Height of the given tree is: "+height);
}
}
|
Javascript
<script>
function rec(i,parent,height)
{
if (parent[i] == -1) {
return 1;
}
if (height[i] != -1) {
return height[i];
}
height[i] = rec(parent[i], parent, height) + 1;
return height[i];
}
function findHeight(parent,n)
{
let res = 0;
let height=new Array(n);
for(let i=0;i<n;i++)
{
height[i]=-1;
}
for (let i = 0; i < n; i++) {
res = Math.max(res, rec(i, parent, height));
}
return res;
}
let parent=[-1, 0, 1, 6, 6, 0, 0, 2, 7];
let n=parent.length;
let height = findHeight(parent, n);
document.write("Height of the given tree is: "+height+"<br>");
</script>
|
Output:
Height of the given tree is: 5
Time complexity :- O(n)
Space complexity :- O(n)
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Last Updated :
08 Mar, 2023
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