Sub-tree with minimum color difference in a 2-coloured tree

A tree with N nodes and N-1 edges is given with 2 different colours for its nodes.
Find the sub-tree with minimum colour difference i.e. abs(1-colour nodes – 2-colour nodes) is minimum.


Input : 
Edges : 1 2
        1 3
        2 4
        3 5
Colours : 1 1 2 2 1 [1-based indexing where 
                    index denotes the node]
Output : 2
Explanation : The sub-tree {1-2} and {1-2-3-5}
have color difference of 2. Sub-tree {1-2} has two
1-colour nodes and zero 2-colour nodes. So, color 
difference is 2. Sub-tree {1-2-3-5} has three 1-colour
nodes and one 2-colour nodes. So color diff = 2.

Method 1 : The problem can be solved by checking every possible sub-tree from every node of the tree. This will take exponential time as we will check for sub-trees from every node.

Method 2 : (Efficient) If we observe, we are solving a portion of the tree several times. This produces recurring sub-problems. We can use Dynamic Programming approach to get the minimum color difference in one traversal. To make things simpler, we can have color values as 1 and -1. Now, if we have a sub-tree with both colored nodes equal, our sum of colors will be 0. To get the minimum difference, we should have maximum negative sum or maximum positive sum.

  • Case 1 When we need to have a sub-tree with maximum sum : We take a node if its value > 0, i.e. sum(parent) += max(0, sum(child))
  • Case 2 When we need to have a sub-tree with minimum sum(or max negative sum) : We take a node if its value < 0, i.e. sum(parent) += min(0, sum(child))

To get the minimum sum, we can interchange the colors of nodes, i.e. -1 becomes 1 and vice-versa.

Below is the C++ implementation :





// CPP code to find the sub-tree with minimum color
// difference in a 2-coloured tree
#include <bits/stdc++.h>
using namespace std;
// Tree traversal to compute minimum difference
void dfs(int node, int parent, vector<int> tree[], 
                    int colour[], int answer[])
    // Initial min difference is the color of node
    answer[node] = colour[node];
    // Traversing its children
    for (auto u : tree[node]) {
        // Not traversing the parent
        if (u == parent)
        dfs(u, node, tree, colour, answer);
        // If the child is adding positively to
        // difference, we include it in the answer
        // Otherwise, we leave the sub-tree and 
        // include 0 (nothing) in the answer
        answer[node] += max(answer[u], 0);
int maxDiff(vector<int> tree[], int colour[], int N)
       int answer[N + 1];
       memset(answer, 0, sizeof(answer));
    // DFS for colour difference : 1colour - 2colour
    dfs(1, 0, tree, colour, answer);
    // Minimum colour difference is maximum answer value
    int high = 0;
    for (int i = 1; i <= N; i++) {
        high = max(high, answer[i]);
        // Clearing the current value
        // to check for colour2 as well
        answer[i] = 0;
    // Interchanging the colours
    for (int i = 1; i <= N; i++) {
        if (colour[i] == -1)
            colour[i] = 1;
            colour[i] = -1;
    // DFS for colour difference : 2colour - 1colour
    dfs(1, 0, tree, colour, answer);
    // Checking if colour2 makes the minimum colour 
    // difference
    for (int i = 1; i < N; i++)
        high = max(high, answer[i]);
    return high;
// Driver code
int main()
    // Nodes
    int N = 5;
    // Adjacency list representation
    vector<int> tree[N + 1];
    // Edges
    // Index represent the colour of that node
    // There is no Node 0, so we start from 
    // index 1 to N
    int colour[] = { 0, 1, 1, -1, -1, 1 };
    // Printing the result
    cout << maxDiff(tree,  colour,  N);
    return 0;




This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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