Prerequisite: Segment tree and depth first search.
In this article, an approach to convert an N-ary rooted tree( a tree with more than 2 children) into a segment tree is discussed which is used to perform a range update queries.
Why do we need a segment tree when we already have an n-ary rooted tree?
Many times, a situation occurs where the same operation has to be performed on multiple nodes and their subtrees along with query operations multiple times.
Let’s say that we have to perform N updates on different subtrees. Every operation can take up to O(N) time as it is an N-ary tree so overall complexity will be O(N^2) which is too slow to process more than 10^3 updates queries. So we have to go the other way around and we will build a segment tree for the same.
Approach: A depth first search is performed to walk through all the nodes and keep the track of the indexes of the subtree of every node in a converted array using two arrays tin and tout(which will be the range to do updates and queries). The DFS will perform a Euler walk. The idea is to create an array and add nodes to it in the order they get visited to the converted array.
Let’s see how the tin and tout arrays help in determining the range in the converted array.
Let N-ary rooted tree be:
real values on nodes: 1 2 2 1 4 3 6 converted arr(indexes): 1 2 3 5 6 7 4 Node 3 has three children 5, 6, 7. Therefore, the range of node 3 is index 3-6. NODE: RANGE(tin-tout) NODE 1: 1 - 7 NODE 2: 2 - 2 NODE 3: 3 - 6 NODE 5: 4 - 4 NODE 6: 5 - 5 NODE 7: 6 - 6 NODE 4: 7 - 7
Here, Node 1 has a range from 1-7 (all nodes) so the update and query will be performed on all the nodes. Leaf nodes like 2 which have no children will only update range 2-2(only itself) this proves that our range arrays tin and tout are correct. Similarly, tin and tout for all the nodes determine the range for query and update in the segment tree.
The following is the implementation of the approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define N 100005 // Keeping the values array indexed by 1. int arr[8] = { 0, 1, 2, 2, 1, 4, 3, 6 }; vector<int> tree[N]; int idx, tin[N], tout[N], converted[N]; // Function to perform DFS in the tree void dfs(ll node, ll parent) { ++idx; converted[idx] = node; // To store starting range of a node tin[node] = idx; for (auto i : tree[node]) { if (i != parent) dfs(i, node); } // To store ending range of a node tout[node] = idx; } // Segment tree ll t[N * 4]; // Build using the converted array indexes. // Here a simple n-ary tree is converted // into a segment tree. // Now O(NlogN) range updates and queries // can be performed. void build(ll node, ll start, ll end) { if (start == end) t[node] = arr[converted[start]]; else { ll mid = (start + end) >> 1; build(2 * node, start, mid); build(2 * node + 1, mid + 1, end); t[node] = t[2 * node] + t[2 * node + 1]; } } // Function to perform update operation // on the tree void update(ll node, ll start, ll end, ll lf, ll rg, ll c) { if (start > end or start > rg or end < lf) return; if (start == end) { t[node] = c; } else { ll mid = (start + end) >> 1; update(2 * node, start, mid, lf, rg, c); update(2 * node + 1, mid + 1, end, lf, rg, c); t[node] = t[2 * node] + t[2 * node + 1]; } } // Function to find the sum at every node ll query(ll node, ll start, ll end, ll lf, ll rg) { if (start > rg or end < lf) return 0; if (lf <= start and end <= rg) { return t[node]; } else { ll ans = 0; ll mid = (start + end) >> 1; ans += query(2 * node, start, mid, lf, rg); ans += query(2 * node + 1, mid + 1, end, lf, rg); return ans; } } // Function to print the tree void printTree(int q, int node, int n) { while (q--) { // Calculating range of node in segment tree ll lf = tin[node]; ll rg = tout[node]; ll res = query(1, 1, n, lf, rg); cout << "sum at node " << node << ": " << res << endl; node++; } } // Driver code int main() { int n = 7; int q = 7; // Creating the tree. tree[1].pb(2); tree[1].pb(3); tree[1].pb(4); tree[3].pb(5); tree[3].pb(6); tree[3].pb(7); // DFS to get converted array. idx = 0; dfs(1, -1); // Build segment tree with converted array. build(1, 1, n); printTree(7, 1, 7); // Updating the value at node 3 int node = 3; ll lf = tin[node]; ll rg = tout[node]; ll value = 4; update(1, 1, n, lf, rg, value); cout << "After Update" << endl; printTree(7, 1, 7); return 0; } |
Java
// Java implementation of the above approach import java.util.*; class GFG { static final int N = 100005; // Keeping the values array indexed by 1. static int arr[] = { 0, 1, 2, 2, 1, 4, 3, 6 }; static Vector<Integer> []tree = new Vector[N]; static int idx; static int []tin = new int[N]; static int []tout = new int[N]; static int []converted = new int[N]; // Function to perform DFS in the tree static void dfs(int node, int parent) { ++idx; converted[idx] = node; // To store starting range of a node tin[node] = idx; for (int i : tree[node]) { if (i != parent) dfs(i, node); } // To store ending range of a node tout[node] = idx; } // Segment tree static int []t = new int[N * 4]; // Build using the converted array indexes. // Here a simple n-ary tree is converted // into a segment tree. // Now O(NlogN) range updates and queries // can be performed. static void build(int node, int start, int end) { if (start == end) t[node] = arr[converted[start]]; else { int mid = (start + end) >> 1; build(2 * node, start, mid); build(2 * node + 1, mid + 1, end); t[node] = t[2 * node] + t[2 * node + 1]; } } // Function to perform update operation // on the tree static void update(int node, int start, int end, int lf, int rg, int c) { if (start > end || start > rg || end < lf) return; if (start == end) { t[node] = c; } else { int mid = (start + end) >> 1; update(2 * node, start, mid, lf, rg, c); update(2 * node + 1, mid + 1, end, lf, rg, c); t[node] = t[2 * node] + t[2 * node + 1]; } } // Function to find the sum at every node static int query(int node, int start, int end, int lf, int rg) { if (start > rg || end < lf) return 0; if (lf <= start && end <= rg) { return t[node]; } else { int ans = 0; int mid = (start + end) >> 1; ans += query(2 * node, start, mid, lf, rg); ans += query(2 * node + 1, mid + 1, end, lf, rg); return ans; } } // Function to print the tree static void printTree(int q, int node, int n) { while (q-- > 0) { // Calculating range of node in segment tree int lf = tin[node]; int rg = tout[node]; int res = query(1, 1, n, lf, rg); System.out.print("sum at node " + node + ": " + res +"\n"); node++; } } // Driver code public static void main(String[] args) { int n = 7; int q = 7; for(int i = 0; i < N; i++) tree[i] = new Vector<Integer>(); // Creating the tree. tree[1].add(2); tree[1].add(3); tree[1].add(4); tree[3].add(5); tree[3].add(6); tree[3].add(7); // DFS to get converted array. idx = 0; dfs(1, -1); // Build segment tree with converted array. build(1, 1, n); printTree(7, 1, 7); // Updating the value at node 3 int node = 3; int lf = tin[node]; int rg = tout[node]; int value = 4; update(1, 1, n, lf, rg, value); System.out.print("After Update" + "\n"); printTree(7, 1, 7); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the above approach N = 100005 # Keeping the values array indexed by 1. arr = [0, 1, 2, 2, 1, 4, 3, 6] tree = [[] for i in range(N)] idx = 0tin = [0]*N tout = [0]*N converted = [0]*N # Function to perform DFS in the tree def dfs(node, parent): global idx idx += 1 converted[idx] = node # To store starting range of a node tin[node] = idx for i in tree[node]: if (i != parent): dfs(i, node) # To store ending range of a node tout[node] = idx # Segment tree t = [0]*(N * 4) # Build using the converted array indexes. # Here a simple n-ary tree is converted # into a segment tree. # Now O(NlogN) range updates and queries # can be performed. def build(node, start, end): if (start == end): t[node] = arr[converted[start]] else: mid = (start + end) >> 1 build(2 * node, start, mid) build(2 * node + 1, mid + 1, end) t[node] = t[2 * node] + t[2 * node + 1] # Function to perform update operation # on the tree def update(node, start, end,lf, rg, c): if (start > end or start > rg or end < lf): return if (start == end): t[node] = c else: mid = (start + end) >> 1 update(2 * node, start, mid, lf, rg, c) update(2 * node + 1, mid + 1, end, lf, rg, c) t[node] = t[2 * node] + t[2 * node + 1] # Function to find the sum at every node def query(node, start, end, lf, rg): if (start > rg or end < lf): return 0 if (lf <= start and end <= rg): return t[node] else: ans = 0 mid = (start + end) >> 1 ans += query(2 * node, start, mid, lf, rg) ans += query(2 * node + 1, mid + 1, end, lf, rg) return ans # Function to print tree def printTree(q, node, n): while (q > 0): # Calculating range of node in segment tree lf = tin[node] rg = tout[node] res = query(1, 1, n, lf, rg) print("sum at node",node,":",res) node += 1 q -= 1 # Driver code if __name__ == '__main__': n = 7 q = 7 # Creating the tree. tree[1].append(2) tree[1].append(3) tree[1].append(4) tree[3].append(5) tree[3].append(6) tree[3].append(7) # DFS to get converted array. idx = 0 dfs(1, -1) # Build segment tree with converted array. build(1, 1, n) printTree(7, 1, 7) # Updating the value at node 3 node = 3 lf = tin[node] rg = tout[node] value = 4 update(1, 1, n, lf, rg, value) print("After Update") printTree(7, 1, 7) # This code is contributed by mohit kumar 29 |
C#
// C# implementation of the above approach using System; using System.Collections.Generic; class GFG { static readonly int N = 100005; // Keeping the values array indexed by 1. static int []arr = { 0, 1, 2, 2, 1, 4, 3, 6 }; static List<int> []tree = new List<int>[N]; static int idx; static int []tin = new int[N]; static int []tout = new int[N]; static int []converted = new int[N]; // Function to perform DFS in the tree static void dfs(int node, int parent) { ++idx; converted[idx] = node; // To store starting range of a node tin[node] = idx; foreach (int i in tree[node]) { if (i != parent) dfs(i, node); } // To store ending range of a node tout[node] = idx; } // Segment tree static int []t = new int[N * 4]; // Build using the converted array indexes. // Here a simple n-ary tree is converted // into a segment tree. // Now O(NlogN) range updates and queries // can be performed. static void build(int node, int start, int end) { if (start == end) t[node] = arr[converted[start]]; else { int mid = (start + end) >> 1; build(2 * node, start, mid); build(2 * node + 1, mid + 1, end); t[node] = t[2 * node] + t[2 * node + 1]; } } // Function to perform update operation // on the tree static void update(int node, int start, int end, int lf, int rg, int c) { if (start > end || start > rg || end < lf) return; if (start == end) { t[node] = c; } else { int mid = (start + end) >> 1; update(2 * node, start, mid, lf, rg, c); update(2 * node + 1, mid + 1, end, lf, rg, c); t[node] = t[2 * node] + t[2 * node + 1]; } } // Function to find the sum at every node static int query(int node, int start, int end, int lf, int rg) { if (start > rg || end < lf) return 0; if (lf <= start && end <= rg) { return t[node]; } else { int ans = 0; int mid = (start + end) >> 1; ans += query(2 * node, start, mid, lf, rg); ans += query(2 * node + 1, mid + 1, end, lf, rg); return ans; } } // Function to print the tree static void printTree(int q, int node, int n) { while (q-- > 0) { // Calculating range of node in segment tree int lf = tin[node]; int rg = tout[node]; int res = query(1, 1, n, lf, rg); Console.Write("sum at node " + node + ": " + res +"\n"); node++; } } // Driver code public static void Main(String[] args) { int n = 7; for(int i = 0; i < N; i++) tree[i] = new List<int>(); // Creating the tree. tree[1].Add(2); tree[1].Add(3); tree[1].Add(4); tree[3].Add(5); tree[3].Add(6); tree[3].Add(7); // DFS to get converted array. idx = 0; dfs(1, -1); // Build segment tree with converted array. build(1, 1, n); printTree(7, 1, 7); // Updating the value at node 3 int node = 3; int lf = tin[node]; int rg = tout[node]; int value = 4; update(1, 1, n, lf, rg, value); Console.Write("After Update" + "\n"); printTree(7, 1, 7); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript implementation of the above approach let N = 100005; // Keeping the values array indexed by 1. let arr = [ 0, 1, 2, 2, 1, 4, 3, 6]; let tree = new Array(N); let idx; let tin = new Array(N); let tout = new Array(N); let converted = new Array(N); // Function to perform DFS in the tree function dfs(node,parent) { ++idx; converted[idx] = node; // To store starting range of a node tin[node] = idx; for (let i=0;i<tree[node].length;i++) { if (tree[node][i] != parent) dfs(tree[node][i], node); } // To store ending range of a node tout[node] = idx; } // Segment tree let t = new Array(N * 4); // Build using the converted array indexes. // Here a simple n-ary tree is converted // into a segment tree. // Now O(NlogN) range updates and queries // can be performed function build(node,start,end) { if (start == end) t[node] = arr[converted[start]]; else { let mid = (start + end) >> 1; build(2 * node, start, mid); build(2 * node + 1, mid + 1, end); t[node] = t[2 * node] + t[2 * node + 1]; } } // Function to perform update operation // on the tree function update(node,start,end,lf,rg,c) { if (start > end || start > rg || end < lf) return; if (start == end) { t[node] = c; } else { let mid = (start + end) >> 1; update(2 * node, start, mid, lf, rg, c); update(2 * node + 1, mid + 1, end, lf, rg, c); t[node] = t[2 * node] + t[2 * node + 1]; } } // Function to find the sum at every node function query(node,start,end,lf,rg) { if (start > rg || end < lf) return 0; if (lf <= start && end <= rg) { return t[node]; } else { let ans = 0; let mid = (start + end) >> 1; ans += query(2 * node, start, mid, lf, rg); ans += query(2 * node + 1, mid + 1, end, lf, rg); return ans; } } // Function to print the tree function printTree(q,node,n) { while (q-- > 0) { // Calculating range of node in segment tree let lf = tin[node]; let rg = tout[node]; let res = query(1, 1, n, lf, rg); document.write("sum at node " + node + ": " + res +"<br>"); node++; } } // Driver code let n = 7; let q = 7; for(let i = 0; i < N; i++) tree[i] = []; // Creating the tree. tree[1].push(2); tree[1].push(3); tree[1].push(4); tree[3].push(5); tree[3].push(6); tree[3].push(7); // DFS to get converted array. idx = 0; dfs(1, -1); // Build segment tree with converted array. build(1, 1, n); printTree(7, 1, 7); // Updating the value at node 3 let node = 3; let lf = tin[node]; let rg = tout[node]; let value = 4; update(1, 1, n, lf, rg, value); document.write("After Update" + "<br>"); printTree(7, 1, 7); // This code is contributed by patel2127 </script> |
Output:
sum at node 1: 19 sum at node 2: 2 sum at node 3: 15 sum at node 4: 1 sum at node 5: 4 sum at node 6: 3 sum at node 7: 6 After Update sum at node 1: 20 sum at node 2: 2 sum at node 3: 16 sum at node 4: 1 sum at node 5: 4 sum at node 6: 4 sum at node 7: 4
Related Topic: Segment Tree
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