Given a tree (not necessarily a binary tree) and a number of queries such that every query takes two nodes of tree as parameters. For every query pair, find if two nodes are on the same path from root to the bottom.
For example, consider the below tree, if given queries are (1, 5), (1, 6) and (2, 6), then answers should be true, true and false respectively.
Note that 1 and 5 lie on same root to leaf path, so do 1 and 6, but 2 and 6 are not on same root to leaf path.
It is obvious that Depth First Search technique is to be used to solve above problem, the main problem is how to respond to multiple queries fast. Here our graph is a tree which may have any number of children. Now DFS in a tree if started from root node proceeds in a depth search manner i.e. Suppose root has three children and those children have only one child with them so if DFS is started then it first visits the first child of root node then will go deep to the child of that node. The situation with a small tree can be shown as follows:
The order of visiting the nodes will be – 1 2 5 3 6 4 7 .
Thus other children nodes are visited later until completely one child is succesfully visited till depth. To simplify this if we assume that we have a watch in our hand and we start walking from root in DFS manner.
Intime – When we visit the node for the first time
Outtime- If we again visit the node later but there is no children unvisited we call it outtime,
Note: Any node in its sub-tree will always have intime < its children (or children of children) because it is always visited first before children (due to DFS) and will have outtime > all nodes in its sub-tree because before noting the outtime it waits for all of its children to be marked visited.
For any two nodes u, v if they are in same path then,
Intime[v] < Intime[u] and Outtime[v] > Outtime[u]
OR
Intime[u] < Intime[v] and Outtime[u ]> Outtime[v]
- If given pair of nodes follows any of the two conditions, then they are on the same root to leaf path.
- Else not on same path (If two nodes are on different paths it means that no one is in subtree of each other).
Pseudo Code
We use a global variable time which will be incremented as dfs for a node begins and will also be incremented after
DFS(v)
increment timer
Intime[v] = timer
mark v as visited
for all u that are children of v
DFS(u)
increment timer
Outtime[v] = timer
end
Time Complexity – O(n) for preprocessing and O(1) per query.
Implementation:
Below is implementation of above pseudo code.
C++
// C++ program to check if given pairs lie on same // path or not. #include<bits/stdc++.h> using namespace std; const int MAX = 100001; // To keep track of visited vertices in DFS bool visit[MAX] = {0}; // To store start and end time of all vertices // during DFS. int intime[MAX]; int outtime[MAX]; // initially timer is zero int timer = 0; // Does DFS of given graph and fills arrays // intime[] and outtime[]. These arrays are used // to answer given queries. void dfs(vector<int> graph[], int v) { visit[v] = true; // Increment the timer as you enter // the recursion for v ++timer; // Upgrade the in time for the vertex intime[v] = timer; vector<int>::iterator it = graph[v].begin(); while (it != graph[v].end()) { if (visit[*it]==false) dfs(graph, *it); it++; } // increment the timer as you exit the // recursion for v ++timer; // upgrade the outtime for that node outtime[v] = timer; } // Returns true if 'u' and 'v' lie on same root to leaf path // else false. bool query(int u, int v) { return ( (intime[u]<intime[v] && outtime[u]>outtime[v]) || (intime[v]<intime[u] && outtime[v]>outtime[u]) ); } // Driver code int main() { // Let us create above shown tree int n = 9; // total number of nodes vector<int> graph[n+1]; graph[1].push_back(2); graph[1].push_back(3); graph[3].push_back(6); graph[2].push_back(4); graph[2].push_back(5); graph[5].push_back(7); graph[5].push_back(8); graph[5].push_back(9); // Start dfs (here root node is 1) dfs(graph, 1); // below are calls for few pairs of nodes query(1, 5)? cout << "Yes
" : cout << "No
"; query(2, 9)? cout << "Yes
" : cout << "No
"; query(2, 6)? cout << "Yes
" : cout << "No
"; return 0; } |
Python3
# Python3 program to check if given # pairs lie on same path or not. # Does DFS of given graph and fills # arrays intime[] and outtime[]. # These arrays are used to answer # given queries. def dfs(graph, v): global intime, outtime, visit, MAX, timer visit[v] = True # Increment the timer as you enter # the recursion for v timer += 1 # Upgrade the in time for the vertex intime[v] = timer it = 0 while it < len(graph[v]): if (visit[graph[v][it]] == False): dfs(graph, graph[v][it]) it += 1 # increment the timer as you # exit the recursion for v timer += 1 # upgrade the outtime for that node outtime[v] = timer # Returns true if 'u' and 'v' lie on # same root to leaf path else false. def query(u, v): global intime, outtime, visit, MAX, timer return ((intime[u] < intime[v] and outtime[u] > outtime[v]) or (intime[v] < intime[u] and outtime[v] > outtime[u]) ) # Driver code MAX = 100001 # To keep track of visited vertices in DFS visit = [0] * MAX # To store start and end time of # all vertices during DFS. intime = [0] * MAXouttime = [0] * MAX # initially timer is zero timer = 0 # Let us create above shown tree n = 9 # total number of nodes graph = [[] for i in range(n+1)] graph[1].append(2) graph[1].append(3) graph[3].append(6) graph[2].append(4) graph[2].append(5) graph[5].append(7) graph[5].append(8) graph[5].append(9) # Start dfs (here root node is 1) dfs(graph, 1) # below are calls for few pairs of nodes print("Yes") if query(1, 5) else print("No") print("Yes") if query(2, 9) else print("No") print("Yes") if query(2, 6) else print("No") # This code is contributed by PranchalK |
Output:
Yes Yes No
Illustration:
From the diagram below to understand more we can have some examples. DFS algorithm as modified above will result in the following intime and outtime for the vertex of the tree as labelled there. Now we will consider all the cases.
Case 1 : Nodes 2 and 4 : Node 2 has intime less than node 4 but since 4 is in its sub tree so it will have a greater exit time than 4 . Thus conditon is valid and both are on same path.
Case 2 : Nodes 7 and 6 : Node 7 has intime less than node 6 but since both nodes are not in each other’s sub tree so their exit time does not follow the required condition.
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