Print all nodes present in the subtree of a given node of a Binary Tree

Last Updated : 01 Feb, 2022

Given two arrays Node_ID[] and Parent_ID[]., construct a binary tree where value of i^thnode is equal to Node_ID[i] and parent of i^th node is Parent_ID[i]. Given a node X, the task is to print node values of the tree rooted at X.

Examples:

Input: Node_ID[]= [11, 48, 100, 5], Parent_ID[] = [48, 0, 5, 48], X = 5
Output: [5, 100]
Explanation:
The tree constructed is as follows:
48
/ \
11 5
/
100
Therefore, subtree of the node 5 contains the nodes {5, 100}.

Input: Node_ID[] = [1, 2, 3], Parent_ID[] = [0, 1, 1], X = 2
Output: [2]

Naive Approach: Follow the steps below to solve the problem

Construct a tree structure from Node_ID[] and Parent_ID[]
Store the nodes with parent X in vector tree
For each node, check if X is an ancestor of that node
If found to be true, store the node in vector tree. Otherwise, continue.
Print the nodes present in vector tree

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
// Function to print nodes
// in the tree rooted at x
void subtreeX(vector<int>& nid,
              vector<int>& pid, int x)
{
    unordered_map<int, int> parent;
 
    vector<int> tree;
 
    // Map every node to its parent
    for (int i = 0; i < nid.size(); i++) {
        parent[nid[i]] = pid[i];
    }
 
    // Subtree with x as root
    tree.push_back(x);
 
    for (int i = 0; i < nid.size(); i++) {
        int k = nid[i];
        int p = k;
 
        // Iterate until k becomes
        // equal to the root
        while (k != 0) {
 
            if (parent[k] == x) {
 
                // x is an ancestor of nid[i]
                tree.push_back(nid[i]);
                break;
            }
 
            k = parent[k];
        }
    }
 
    // Print elements in the subtree
    for (int node : tree)
        cout << node << " ";
}
 
// Driver Code
int main()
{
    vector<int> nid = { 11, 48, 100, 5 };
    vector<int> pid = { 48, 0, 5, 48 };
    int x = 5;
 
    // Function call to print nodes
    // in the tree rooted at x
    subtreeX(nid, pid, x);
 
    return 0;
}

Java

import java.util.*;
class GFG 
{
 
    // Function to print nodes
    // in the tree rooted at x
    static void subtreeX(int[] nid, int[] pid, int x) 
    {
        HashMap<Integer, Integer> parent 
        = new HashMap<Integer, Integer>();
 
        List<Integer> tree = new LinkedList<>();
 
        // Map every node to its parent
        for (int i = 0; i < nid.length; i++)
        {
            parent.put(nid[i], pid[i]);
        }
 
        // Subtree with x as root
        tree.add(x);
        for (int i = 0; i < nid.length; i++)
        {
            int k = nid[i];
            int p = k;
 
            // Iterate until k becomes
            // equal to the root
            while (k != 0) 
            {
                 
                if (parent.containsKey(k) && parent.get(k) == x) 
                {
                   
                    // x is an ancestor of nid[i]
                    tree.add(nid[i]);
                    break;
                }
                k = parent.containsKey(k) ? parent.get(k) : -1;
            }
        }
 
        // Print elements in the subtree
        for (int node : tree)
            System.out.print(node + " ");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] nid = { 11, 48, 100, 5 };
        int[] pid = { 48, 0, 5, 48 };
        int x = 5;
 
        // Function call to print nodes
        // in the tree rooted at x
        subtreeX(nid, pid, x);
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Function to print nodes
# in the tree rooted at x
def subtreeX(nid, pid, x):
    parent = {}
    tree = []
 
    # Map every node to its parent
    for i in range(len(nid)):
        parent[nid[i]] = pid[i]
 
    # Subtree with x as root
    tree.append(x)
    for i in range(len(nid)):
        k = nid[i]
        p = k
 
        # Iterate until k becomes
        # equal to the root
        while (k != 0):
            if (parent[k] == x):
 
                # x is an ancestor of nid[i]
                tree.append(nid[i])
                break
            k = parent[k]
 
    # Print elements in the subtree
    for node in tree:
        print(node, end = " ")
 
# Driver Code
if __name__ == '__main__':
    nid = [11, 48, 100, 5]
    pid = [48, 0, 5, 48 ]
    x = 5
 
    # Function call to print nodes
    # in the tree rooted at x
    subtreeX(nid, pid, x)
 
    # This code is contributed by mohit kumar 29.

C#

using System;
using System.Collections.Generic;
public class GFG 
{
 
  // Function to print nodes
  // in the tree rooted at x
  static void subtreeX(int[] nid, int[] pid, int x) 
  {
    Dictionary<int, int> parent 
      = new Dictionary<int, int>();
    List<int> tree = new List<int>();
 
    // Map every node to its parent
    for (int i = 0; i < nid.Length; i++)
    {
      parent.Add(nid[i], pid[i]);
    }
 
    // Subtree with x as root
    tree.Add(x);
    for (int i = 0; i < nid.Length; i++)
    {
      int k = nid[i];
      int p = k;
 
      // Iterate until k becomes
      // equal to the root
      while (k != 0) 
      {       
        if (parent.ContainsKey(k) && parent[k] == x) 
        {
 
          // x is an ancestor of nid[i]
          tree.Add(nid[i]);
          break;
        }
        k = parent.ContainsKey(k) ? parent[k] : -1;
      }
    }
 
    // Print elements in the subtree
    foreach (int node in tree)
      Console.Write(node + " ");
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int[] nid = { 11, 48, 100, 5 };
    int[] pid = { 48, 0, 5, 48 };
    int x = 5;
 
    // Function call to print nodes
    // in the tree rooted at x
    subtreeX(nid, pid, x);
  }
}
 
// This code is contributed by shikhasingrajput

Javascript

<script>
 
// Function to print nodes
// in the tree rooted at x
function subtreeX(nid, pid, x) 
{
  var parent = new Map();
 
  var tree = [];
  // Map every node to its parent
  for (var i = 0; i < nid.length; i++)
  {
    parent.set(nid[i], pid[i]);
  }
  // Subtree with x as root
  tree.push(x);
  for (var i = 0; i < nid.length; i++)
  {
    var k = nid[i];
    var p = k;
    // Iterate until k becomes
    // equal to the root
    while (k != 0) 
    {       
      if (parent.has(k) && parent.get(k) == x) 
      {
        // x is an ancestor of nid[i]
        tree.push(nid[i]);
        break;
      }
      k = parent.has(k) ? parent.get(k) : -1;
    }
  }
  // Print elements in the subtree
  for(var node of tree)
    document.write(node + " ");
}
// Driver Code
var nid = [11, 48, 100, 5];
var pid = [48, 0, 5, 48];
var x = 5;
// Function call to print nodes
// in the tree rooted at x
subtreeX(nid, pid, x);
 
 
 
</script>

Output:

5 100

Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach : Follow the steps below to optimize the above approach:

Construct a tree structure from Node_ID[] and Parent_ID[]
Perform DFS from node X.
Store the nodes in a vector tree
Print the nodes present in the vector tree

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
// DFS to traverse subtree rooted  at x
void dfs(int x, vector<int>& tree,
         map<int, vector<int> >& child)
{
    // Push x into the vector
    tree.push_back(x);
 
    // Check if x is a leaf node
    if (child.find(x) != child.end()) {
 
        // Recursively call dfs
        // for children of x
        for (int next : child[x]) {
            dfs(next, tree, child);
        }
    }
}
 
// Function to print nodes
// in the tree rooted at x
void SubtreeX(vector<int>& nid,
              vector<int>& pid, int x)
{
    int n = nid.size();
    map<int, vector<int> > child;
 
    // adding edges in a tree
    for (int i = 0; i < n; i++) {
 
        if (child.find(pid[i])
            == child.end()) {
 
            // Initialize adjacency list
            child[pid[i]] = vector<int>();
        }
 
        child[pid[i]].push_back(nid[i]);
    }
 
    // Stores nodes in the subtree
    vector<int> tree;
 
    // Perform DFS from node x
    dfs(x, tree, child);
 
    for (int node : tree) {
        cout << node << " ";
    }
}
 
// Driver Code
int main()
{
    vector<int> nid = { 11, 48, 100, 5 };
    vector<int> pid = { 48, 0, 5, 48 };
    int x = 5;
 
    // Function call to print nodes
    // in the tree rooted at x
    SubtreeX(nid, pid, x);
 
    return 0;
}

Java

import java.io.*;
import java.util.*;
 
class GFG{
     
// DFS to traverse subtree rooted  at x
static void dfs(int x, Vector<Integer> tree,
          Map<Integer, Vector<Integer>> child)
{
     
    // Push x into the vector
    tree.add(x);
  
    // Check if x is a leaf node
    if (child.containsKey(x)) 
    {
         
        // Recursively call dfs
        // for children of x
        for(int next : child.get(x))
        {
            dfs(next, tree, child);
        }
    }
}
  
// Function to print nodes
// in the tree rooted at x
static void SubtreeX(Vector<Integer> nid,
                     Vector<Integer> pid, int x)
{
    int n = nid.size();
    Map<Integer, Vector<Integer>> child = new HashMap<>();
  
    // Adding edges in a tree
    for(int i = 0; i < n; i++) 
    {
        if (!child.containsKey(pid.get(i))) 
        {
             
            // Initialize adjacency list
            child.put(pid.get(i), new Vector<Integer>());
        }
        child.get(pid.get(i)).add(nid.get(i));
    }
  
    // Stores nodes in the subtree
    Vector<Integer> tree = new Vector<Integer>();
  
    // Perform DFS from node x
    dfs(x, tree, child);
  
    for(int node : tree) 
    {
        System.out.print(node + " ");
    }
}
  
// Driver Code
public static void main (String[] args)
{
    Vector<Integer> nid = new Vector<Integer>(
        Arrays.asList(11, 48, 100, 5));
    Vector<Integer> pid = new Vector<Integer>(
        Arrays.asList(48, 0, 5, 48));
     
    int x = 5;
 
    // Function call to print nodes
    // in the tree rooted at x
    SubtreeX(nid, pid, x);
}
}
 
// This code is contributed by rag2127

Python3

# DFS to traverse subtree rooted  at x
def dfs(x, tree, child):
   
    # Push x into the vector
    tree.append(x)
     
    # Check if x is a leaf node
    if x in child:
       
        # Recursively call dfs
        # for children of x
        for nextt in child[x]:
            dfs(nextt, tree, child)
 
# Function to print nodes
# in the tree rooted at x
def SubtreeX(nid,pid,x):
    n = len(nid)
    child = {}
     
    #  Adding edges in a tree   
    for i in range(n):
        if pid[i] not in child:
           
            # Initialize adjacency list
            child[pid[i]] = []
        child[pid[i]].append(nid[i])
     
    # Stores nodes in the subtree
    tree = []
     
    # Perform DFS from node x
    dfs(x, tree, child)
    print(*tree)
     
# Driver Code
nid = [11, 48, 100, 5]
pid = [48, 0, 5, 48]
x = 5
 
# Function call to print nodes
# in the tree rooted at x
SubtreeX(nid, pid, x)
 
# This code is contributed by avanitrachhadiya2155

C#

using System;
using System.Collections.Generic;
class GFG {
     
    // DFS to traverse subtree rooted  at x
    static void dfs(int x, List<int> tree,
              Dictionary<int, List<int>> child)
    {
          
        // Push x into the vector
        tree.Add(x);
       
        // Check if x is a leaf node
        if (child.ContainsKey(x))
        {
              
            // Recursively call dfs
            // for children of x
            foreach(int next in child[x])
            {
                dfs(next, tree, child);
            }
        }
    }
       
    // Function to print nodes
    // in the tree rooted at x
    static void SubtreeX(List<int> nid, List<int> pid, int x)
    {
        int n = nid.Count;
        Dictionary<int, List<int>> child = new Dictionary<int, List<int>>();
       
        // Adding edges in a tree
        for(int i = 0; i < n; i++)
        {
            if (!child.ContainsKey(pid[i]))
            {
                  
                // Initialize adjacency list
                child[pid[i]] = new List<int>();
            }
            child[pid[i]].Add(nid[i]);
        }
       
        // Stores nodes in the subtree
        List<int> tree = new List<int>();
       
        // Perform DFS from node x
        dfs(x, tree, child);
       
        foreach(int node in tree)
        {
            Console.Write(node + " ");
        }
    }
 
  // Driver code
  static void Main() {
    List<int> nid = new List<int>{11, 48, 100, 5};
    List<int> pid = new List<int>{48, 0, 5, 48};
      
    int x = 5;
  
    // Function call to print nodes
    // in the tree rooted at x
    SubtreeX(nid, pid, x);
  }
}
 
// This code is contributed by decode2207.

Javascript

<script>
 
// DFS to traverse subtree rooted  at x
function dfs(x,tree,child)
{
    // Push x into the vector
    tree.push(x);
       
    // Check if x is a leaf node
    if (child.has(x))
    {
          
        // Recursively call dfs
        // for children of x
        for(let next=0;next< child.get(x).length;next++)
        {
            dfs(child.get(x)[next], tree, child);
        }
    }
}
 
// Function to print nodes
// in the tree rooted at x
function SubtreeX(nid,pid,x)
{
    let n = nid.length;
    let child = new Map();
   
    // Adding edges in a tree
    for(let i = 0; i < n; i++)
    {
        if (!child.has(pid[i]))
        {
              
            // Initialize adjacency list
            child.set(pid[i], []);
        }
        child.get(pid[i]).push(nid[i]);
    }
   
    // Stores nodes in the subtree
    let tree = [];
   
    // Perform DFS from node x
    dfs(x, tree, child);
   
    for(let node=0;node< tree.length;node++)
    {
        document.write(tree[node] + " ");
    }
}
 
// Driver Code
let nid = [11, 48, 100, 5];
let pid = [48, 0, 5, 48];
 
let x = 5;
 
// Function call to print nodes
// in the tree rooted at x
SubtreeX(nid, pid, x);
 
 
// This code is contributed by unknown2108
</script>

Output:

5 100

Time Complexity: O(N)
Auxiliary Space: O(N)

Suggest improvement

Count paths of Tree such that starting and ending node values are same

Sum of all vertical levels of a Binary Tree

Share your thoughts in the comments

Print all nodes present in the subtree of a given node of a Binary Tree

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