# Queries for M-th node in the DFS of subtree

Given a tree of N nodes and N-1 edges. Also given an integer M and a node, the task is to print the M-th node in the DFS of the subtree of a given node for multiple queries.

Note: M will not be greater than the number of nodes in the subtree of the given node. Input: M = 3, node = 1
Output: 4
In the above example if 1 is given as the node, then the DFS of subtree will be 1 2 4 6 7 5 3, hence if M is 3, then the 3rd node is 4

Input: M = 4, node = 2
Output: 7
If 2 is given as the node, then the DFS of the subtree will be 2 4 6 7 5., hence if M is 4 then the 4th node is 7.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Call DFS function to generate the DFS of the complete tree.
• Use an under[] array to store the height of the subtree under the given node including the node.
• In the DFS function, keep incrementing the size of subtree on every recursive call.
• Mark the node index in the DFS of complete using hashing.
• Let index of given node in the DFS of the tree be ind, then the M-th node will be at index ind + M -1 as the DFS of a subtree of a node will always be a contiguous subarray starting from the node.

Below is the implementation of the above approach.

## C++

 `// C++ program for Queries ` `// for DFS of subtree of a node in a tree ` `#include ` `using` `namespace` `std; ` `const` `int` `N = 100000; ` ` `  `// Adjaceny list to store the ` `// tree nodes connection ` `vector<``int``> v[N]; ` ` `  `// stores the index of node in DFS ` `unordered_map<``int``, ``int``> mp; ` ` `  `// stores the index of node in ` `// original node ` `vector<``int``> a; ` ` `  `// Function to call DFS and count nodes ` `// under that subtree ` `void` `dfs(``int` `under[], ``int` `child, ``int` `parent) ` `{ ` ` `  `    ``// stores the DFS of tree ` `    ``a.push_back(child); ` ` `  `    ``// hieght of subtree ` `    ``under[child] = 1; ` ` `  `    ``// iterate for children ` `    ``for` `(``auto` `it : v[child]) { ` ` `  `        ``// if not equal to parent ` `        ``// so that it does not traverse back ` `        ``if` `(it != parent) { ` ` `  `            ``// call DFS for subtree ` `            ``dfs(under, it, child); ` ` `  `            ``// add the height ` `            ``under[child] += under[it]; ` `        ``} ` `    ``} ` `} ` ` `  `// Function to return the DFS of subtree of node ` `int` `printnodeDFSofSubtree(``int` `node, ``int` `under[], ``int` `m) ` `{ ` `    ``// index of node in the original DFS ` `    ``int` `ind = mp[node]; ` ` `  `    ``// height of subtree of node ` `    ``return` `a[ind + m - 1]; ` `} ` ` `  `// Function to add edges to a tree ` `void` `addEdge(``int` `x, ``int` `y) ` `{ ` `    ``v[x].push_back(y); ` `    ``v[y].push_back(x); ` `} ` ` `  `// Marks the index of node in original DFS ` `void` `markIndexDfs() ` `{ ` `    ``int` `size = a.size(); ` ` `  `    ``// marks the index ` `    ``for` `(``int` `i = 0; i < size; i++) { ` `        ``mp[a[i]] = i; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 7; ` ` `  `    ``// add edges of a tree ` `    ``addEdge(1, 2); ` `    ``addEdge(1, 3); ` `    ``addEdge(2, 4); ` `    ``addEdge(2, 5); ` `    ``addEdge(4, 6); ` `    ``addEdge(4, 7); ` ` `  `    ``// array to store the height of subtree ` `    ``// of every node in a tree ` `    ``int` `under[n + 1]; ` ` `  `    ``// Call the function DFS to generate the DFS ` `    ``dfs(under, 1, 0); ` ` `  `    ``// Function call to mark the index of node ` `    ``markIndexDfs(); ` ` `  `    ``int` `m = 3; ` ` `  `    ``// Query 1 ` `    ``cout << printnodeDFSofSubtree(1, under, m) << endl; ` ` `  `    ``// Query 2 ` `    ``m = 4; ` `    ``cout << printnodeDFSofSubtree(2, under, m); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 program for Queries  ` `# for DFS of subtree of a node in a tree  ` `N ``=` `100000` ` `  `# Adjaceny list to store the  ` `# tree nodes connection  ` `v ``=` `[[]``for` `i ``in` `range``(N)] ` ` `  `# stores the index of node in DFS  ` `mp ``=` `{} ` ` `  `# stores the index of node in  ` `# original node  ` `a ``=` `[] ` ` `  `# Function to call DFS and count nodes  ` `# under that subtree  ` `def` `dfs(under, child, parent): ` `     `  `    ``# stores the DFS of tree  ` `    ``a.append(child)  ` `     `  `    ``# hieght of subtree  ` `    ``under[child] ``=` `1` `     `  `    ``# iterate for children  ` `    ``for` `it ``in` `v[child]: ` `         `  `        ``# if not equal to parent  ` `        ``# so that it does not traverse back  ` `        ``if` `(it !``=` `parent): ` `             `  `            ``# call DFS for subtree  ` `            ``dfs(under, it, child)  ` `             `  `            ``# add the height  ` `            ``under[child] ``+``=` `under[it]  ` `             `  `# Function to return the DFS of subtree of node  ` `def` `printnodeDFSofSubtree(node, under, m): ` `     `  `    ``# index of node in the original DFS  ` `    ``ind ``=` `mp[node]  ` `     `  `    ``# height of subtree of node  ` `    ``return` `a[ind ``+` `m ``-` `1``]  ` `     `  `# Function to add edges to a tree  ` `def` `addEdge(x, y): ` `    ``v[x].append(y)  ` `    ``v[y].append(x)  ` ` `  `# Marks the index of node in original DFS  ` `def` `markIndexDfs(): ` `     `  `    ``size ``=` `len``(a) ` `     `  `    ``# marks the index  ` `    ``for` `i ``in` `range``(size): ` `        ``mp[a[i]] ``=` `i  ` `     `  `# Driver Code  ` ` `  `n ``=` `7` ` `  `# add edges of a tree  ` `addEdge(``1``, ``2``)  ` `addEdge(``1``, ``3``)  ` `addEdge(``2``, ``4``)  ` `addEdge(``2``, ``5``)  ` `addEdge(``4``, ``6``)  ` `addEdge(``4``, ``7``)  ` ` `  `# array to store the height of subtree  ` `# of every node in a tree  ` `under ``=` `[``0``]``*``(n ``+` `1``) ` ` `  `# Call the function DFS to generate the DFS  ` `dfs(under, ``1``, ``0``)  ` ` `  `# Function call to mark the index of node  ` `markIndexDfs()  ` ` `  `m ``=` `3` ` `  `# Query 1  ` `print``(printnodeDFSofSubtree(``1``, under, m)) ` ` `  `# Query 2  ` `m ``=` `4` `print``(printnodeDFSofSubtree(``2``, under, m)) ` ` `  `# This code is contributed by SHUBHAMSINGH10 `

Output:

```4
7```

Time Complexity: O(1), for processing each query.
Auxiliary Space: O(N)

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.