# Construct the Rooted tree by using start and finish time of its DFS traversal

• Last Updated : 23 Jun, 2021

Given start and finish times of DFS traversal of N vertices that are available in a Rooted tree, the task is to construct the tree (Print the Parent of each node).
Parent of the root node is 0.
Examples:

```Input: Start[] = {2, 4, 1, 0, 3}, End[] = {3, 5, 4, 5, 4}
Output: 3 4 4 0 3
Given Tree is -:
4(0, 5)
/   \
(1, 4)3     2(4, 5)
/  \
(2, 3)1    5(3, 4)
The root will always have start time = 0
processing a node takes 1 unit time but backtracking
does not consume time, so the finishing time
of two nodes can be the same.

Input: Start[] = {4, 3, 2, 1, 0}, End[] = {5, 5, 3, 3, 5}
Output: 2 5 4 5 0```

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Approach:

• Root of the tree is the vertex whose starting time is zero.
• Now, it is sufficient to find the descendants of a vertex, this way we can find the parent of every vertex.
• Define Identity[i] as the index of the vertex with starting equal to i.
• As Start[v] and End[v] are starting and ending time of vertex v.The first child of v is Identity[Start[v]+1] and
the (i+1)th is Identity[End[chv[i]]] where chv[i] is the ith child of v.
• Traverse down in DFS manner and update the parent of each node.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `int` `N;` `// Function to find the parent of each node.``vector<``int``> Restore_Tree(``int` `Start[], ``int` `End[])``{` `    ``// Storing index of vertex with starting``    ``// time Equal to i``    ``vector<``int``> Identity(N,0);`  `    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``Identity[Start[i]] = i;``    ``}` `    ``// Parent array``    ``vector<``int``> parent(N,-1);``    ``int` `curr_parent = Identity;` `    ``for` `(``int` `j = 1; j < N; j++)``    ``{` `        ``// Find the vertex with starting time j``        ``int` `child = Identity[j];` `        ``// If end time of this child is greater than``        ``// (start time + 1), then we traverse down and``        ``// store curr_parent as the parent of child``        ``if` `(End[child] - j > 1)``        ``{``            ``parent[child] = curr_parent;``            ``curr_parent = child;``        ``}` `        ``// Find the parent of current vertex``        ``// over iterating on the finish time``        ``else``            ``parent[child] = curr_parent;` `        ``// Backtracking takes zero time``        ``while` `(End[child]== End[parent[child]])``        ``{``            ``child = parent[child];``            ``curr_parent = parent[child];``            ``if` `(curr_parent == Identity)``                ``break``;``        ``}``    ``}``    ``for` `(``int` `i = 0; i < N; i++)``        ``parent[i] += 1;` `    ``// Return the parent array``    ``return` `parent;``}` `// Driver Code``int` `main()``{``    ``N = 5;` `    ``// Start and End time of DFS``    ``int` `Start[] = {2, 4, 1, 0, 3};``    ``int` `End[] = {3, 5, 4, 5, 4};``    ``vector<``int``> a = Restore_Tree(Start, End);` `    ``for``(``int` `ans:a)``        ``cout << ans << ``" "``;` `    ``return` `0;``}` `// This code is contributed by mohit kumar 29`

## Java

 `// Java implementation of above approach``import` `java.util.Arrays;` `class` `GFG``{``    ` `static` `int` `N = ``5``;` `// Function to find the parent of each node.``static` `int``[] Restore_Tree(``int` `[]S, ``int` `[]End)``{` `    ``// Storing index of vertex with starting``    ``// time Equal to i``    ``int` `[]Identity = ``new` `int``[N];` `    ``for``(``int` `i = ``0``; i < N; i++)``        ``Identity[S[i]] = i;` `    ``// Parent array``    ``int` `[]parent = ``new` `int``[N];``    ``Arrays.fill(parent,-``1``);``    ``int` `curr_parent = Identity[``0``];``    ` `    ``for``(``int` `j = ``1``; j < N; j++)``    ``{` `        ``// Find the vertex with starting time j``        ``int` `child = Identity[j];` `        ``// If end time of this child is greater than``        ``// (start time + 1), then we traverse down and``        ``// store curr_parent as the parent of child``        ``if``(End[child] - j > ``1``)``        ``{``            ``parent[child] = curr_parent;``            ``curr_parent = child;``        ``}``        ` `        ``// Find the parent of current vertex``        ``// over iterating on the finish time``        ``else``{``            ``parent[child] = curr_parent;` `            ``// Backtracking takes zero time``            ``while``(parent[child]>-``1` `&& End[child] == End[parent[child]])``            ``{``                ``child = parent[child];``                ``curr_parent = parent[child];``                ``if``(curr_parent == Identity[``0``])``                    ``break``;``            ``}``        ``}``    ``}``    ``for``(``int` `i = ``0``; i < N; i++)``        ``parent[i] += ``1``;` `    ``// Return the parent array``    ``return` `parent;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``// Start and End time of DFS``    ``int` `[]Start = {``2``, ``4``, ``1``, ``0``, ``3``};``    ``int` `[]End = {``3``, ``5``, ``4``, ``5``, ``4``};``    ``int` `ans[] =Restore_Tree(Start, End);``    ``for``(``int` `a:ans)``        ``System.out.print(a + ``" "``);``}``}` `// This code has been contributed by 29AjayKumar`

## Python

 `# Python implementation of the above approach`` ` `# Function to find the parent of each node.``def` `Restore_Tree(S, E):`` ` `    ``# Storing index of vertex with starting``    ``# time Equal to i``    ``Identity ``=` `N``*``[``0``] ` `    ``for` `i ``in` `range``(N):``        ``Identity[Start[i]] ``=` `i`` ` `    ``# Parent array``    ``parent ``=` `N``*``[``-``1``]``    ``curr_parent ``=` `Identity[``0``]``    ` `    ``for` `j ``in` `range``(``1``, N):` `        ``# Find the vertex with starting time j``        ``child ``=` `Identity[j]` `        ``# If end time of this child is greater than``        ``# (start time + 1), then we traverse down and``        ``# store curr_parent as the parent of child``        ``if` `End[child] ``-` `j > ``1``:``            ``parent[child] ``=` `curr_parent``            ``curr_parent ``=` `child` `        ``# Find the parent of current vertex``        ``# over iterating on the finish time``        ``else``:    ``            ``parent[child] ``=` `curr_parent` `            ``# Backtracking takes zero time``            ``while` `End[child]``=``=` `End[parent[child]]:``                ``child ``=` `parent[child]``                ``curr_parent ``=` `parent[child]``                ``if` `curr_parent ``=``=` `Identity[``0``]:``                    ``break``    ``for` `i ``in` `range``(N):``        ``parent[i]``+``=` `1`` ` `    ``# Return the parent array``    ``return` `parent`` ` `# Driver Code``if` `__name__``=``=``"__main__"``:``    ``N ``=` `5`` ` `    ``# Start and End time of DFS``    ``Start ``=` `[``2``, ``4``, ``1``, ``0``, ``3``]``    ``End ``=` `[``3``, ``5``, ``4``, ``5``, ``4``]``    ``print``(``*``Restore_Tree(Start, End))`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{``    ` `static` `int` `N = 5;` `// Function to find the parent of each node.``static` `int``[] Restore_Tree(``int` `[]S, ``int` `[]End)``{` `    ``// Storing index of vertex with starting``    ``// time Equal to i``    ``int` `[]Identity = ``new` `int``[N];` `    ``for``(``int` `i = 0; i < N; i++)``        ``Identity[S[i]] = i;` `    ``// Parent array``    ``int` `[]parent = ``new` `int``[N];``    ``for``(``int` `i = 0; i < N; i++)``        ``parent[i]=-1;``    ``int` `curr_parent = Identity;``    ` `    ``for``(``int` `j = 1; j < N; j++)``    ``{` `        ``// Find the vertex with starting time j``        ``int` `child = Identity[j];` `        ``// If end time of this child is greater than``        ``// (start time + 1), then we traverse down and``        ``// store curr_parent as the parent of child``        ``if``(End[child] - j > 1)``        ``{``            ``parent[child] = curr_parent;``            ``curr_parent = child;``        ``}``        ` `        ``// Find the parent of current vertex``        ``// over iterating on the finish time``        ``else``        ``{``            ``parent[child] = curr_parent;` `            ``// Backtracking takes zero time``            ``while``(parent[child]>-1 && End[child] == End[parent[child]])``            ``{``                ``child = parent[child];``                ``curr_parent = parent[child];``                ``if``(curr_parent == Identity)``                    ``break``;``            ``}``        ``}``    ``}``    ``for``(``int` `i = 0; i < N; i++)``        ``parent[i] += 1;` `    ``// Return the parent array``    ``return` `parent;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``// Start and End time of DFS``    ``int` `[]Start = {2, 4, 1, 0, 3};``    ``int` `[]End = {3, 5, 4, 5, 4};``    ``int` `[]ans =Restore_Tree(Start, End);``    ``foreach``(``int` `a ``in` `ans)``        ``Console.Write(a + ``" "``);``}``}` `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``
Output:
`3 4 4 0 3`

Time Complexity : O(N)
where N is the number of nodes in the tree.

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