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Introduction to Dynamic Programming on Trees

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Dynamic Programming(DP) is a technique to solve problems by breaking them down into overlapping sub-problems which follows the optimal substructure. There are various problems using DP like subset sum, knapsack, coin change etc. DP can also be applied on trees to solve some specific problems.
Pre-requisite: DFS
Given a tree with N nodes and N-1 edges, calculate the maximum sum of the node values from root to any of the leaves without re-visiting any node. 
 

Given above is a diagram of a tree with N=14 nodes and N-1=13 edges. The values at node being 3, 2, 1, 10, 1, 3, 9, 1, 5, 3, 4, 5, 9 and 8 respectively for nodes 1, 2, 3, 4….14. 
The diagram below shows all the paths from root to leaves : 
 

All the paths are marked by different colors : 
Path 1(red, 3-2-1-4) : sum of all node values = 10 
Path 2(orange, 3-2-1-5) : sum of all node values = 11 
Path 3(yellow, 3-2-3) : sum of all node values = 8 
Path 4(green, 3-1-9-9) : sum of all node values = 22 
Path 5(violet, 3-1-9-8) : sum of all node values = 21 
Path 6(pink, 3-10-1) : sum of all node values = 14 
Path 7(blue, 3-10-5) : sum of all node values = 18 
Path 8(brown, 3-10-3) : sum of all node values = 16 
The answer is 22, as Path 4 has the maximum sum of values of nodes in its path from a root to leaves. 

The greedy approach fails in this case. Starting from the root and take 3 from the first level, 10 from the next level and 5 from the third level greedily. Result is path-7 if after following the greedy approach, hence do not apply greedy approach over here. 
The problem can be solved using Dynamic Programming on trees. Start memoizing from the leaves and add the maximum of leaves to the root of every sub-tree. At the last step, there will be root and the sub-tree under it, adding the value at node and maximum of sub-tree will give us the maximum sum of the node values from root to any of the leaves.
 

The diagram above shows how to start from the leaves and add the maximum of leaves of a sub-tree to its root. Move upward and repeat the same procedure of storing the maximum of every sub-tree leaves and adding it to its root. In this example, the maximum of node 11 and 12 is taken to count and then added to node 5 (In this sub-tree, 5 is the root and 11, 12 are its leaves). Similarly, the maximum of node 13 and 14 is taken to count and then added to node 7. Repeat the steps for every sub-tree till we reach the node. 

Let DPi be the maximum summation of node values in the path between i and any of its leaves moving downwards. Traverse the tree using DFS traversal. Store the maximum of all the leaves of the sub-tree, and add it to the root of the sub-tree. At the end, DP1 will have the maximum sum of the node values from root to any of the leaves without re-visiting any node. 
Below is the implementation of the above idea : 
 

C++




// C++ code to find the maximum path sum
#include <bits/stdc++.h>
using namespace std;
  
vector<int> dp;
  
// function for dfs traversal and to store the
// maximum value in dp[] for every node till the leaves
void dfs(int a[], vector<int> v[], int u, int parent)
{
    // initially dp[u] is always a[u]
    dp[u] = a[u - 1];
  
    // stores the maximum value from nodes
    int maximum = 0;
  
    // traverse the tree
    for (int child : v[u]) {
  
        // if child is parent, then we continue
        // without recursing further
        if (child == parent)
            continue;
  
        // call dfs for further traversal
        dfs(a, v, child, u);
  
        // store the maximum of previous visited node
        // and present visited node
        maximum = max(maximum, dp[child]);
    }
  
    // add the maximum value returned to the parent node
    dp[u] += maximum;
}
  
// function that returns the maximum value
int maximumValue(int a[], vector<int> v[])
{
    dfs(a, v, 1, 0);
    return dp[1];
}
  
// Driver Code
int main()
{
    // number of nodes
    int n = 14;
  
    // adjacency list
    vector<int> v[n + 1];
  
    // create undirected edges
    // initialize the tree given in the diagram
    v[1].push_back(2), v[2].push_back(1);
    v[1].push_back(3), v[3].push_back(1);
    v[1].push_back(4), v[4].push_back(1);
    v[2].push_back(5), v[5].push_back(2);
    v[2].push_back(6), v[6].push_back(2);
    v[3].push_back(7), v[7].push_back(3);
    v[4].push_back(8), v[8].push_back(4);
    v[4].push_back(9), v[9].push_back(4);
    v[4].push_back(10), v[10].push_back(4);
    v[5].push_back(11), v[11].push_back(5);
    v[5].push_back(12), v[12].push_back(5);
    v[7].push_back(13), v[13].push_back(7);
    v[7].push_back(14), v[14].push_back(7);
  
    // values of node 1, 2, 3....14
    int a[] = { 3, 2, 1, 10, 1, 3, 9, 1, 5, 3, 4, 5, 9, 8 };
    // initialise dp
      dp = vector<int>(n+1,0);
    // function call
    cout << maximumValue(a, v);
  
    return 0;
}


Java




// Java code to find the maximum path sum 
import java.util.Vector;
  
class GFG
{
    static int[] dp = new int[100];
  
    // function for dfs traversal and to 
    // store the maximum value in dp[] 
    // for every node till the leaves
    static void dfs(int[] a, Vector<Integer>[] v, 
                    int u, int parent) 
    {
  
        // initially dp[u] is always a[u]
        dp[u] = a[u - 1];
  
        // stores the maximum value from nodes
        int maximum = 0;
  
        // traverse the tree
        for (int child : v[u]) 
        {
  
            // if child is parent, then we continue
            // without recursing further
            if (child == parent)
                continue;
  
            // call dfs for further traversal
            dfs(a, v, child, u);
  
            // store the maximum of previous visited 
            // node and present visited node
            maximum = Math.max(maximum, dp[child]);
        }
  
        // add the maximum value returned
        // to the parent node
        dp[u] += maximum;
    }
  
    // function that returns the maximum value
    static int maximumValue(int[] a, 
                            Vector<Integer>[] v)
    {
        dfs(a, v, 1, 0);
        return dp[1];
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
  
        // Driver Code
        int n = 14;
  
        // adjacency list
        @SuppressWarnings("unchecked")
        Vector<Integer>[] v = new Vector[n + 1];
  
        for (int i = 0; i < v.length; i++)
            v[i] = new Vector<>();
  
        // create undirected edges
        // initialize the tree given in the diagram
        v[1].add(2); v[2].add(1);
        v[1].add(3); v[3].add(1);
        v[1].add(4); v[4].add(1);
        v[2].add(5); v[5].add(2);
        v[2].add(6); v[6].add(2);
        v[3].add(7); v[7].add(3);
        v[4].add(8); v[8].add(4);
        v[4].add(9); v[9].add(4);
        v[4].add(10); v[10].add(4);
        v[5].add(11); v[11].add(5);
        v[5].add(12); v[12].add(5);
        v[7].add(13); v[13].add(7);
        v[7].add(14); v[14].add(7);
  
        // values of node 1, 2, 3....14
        int a[] = { 3, 2, 1, 10, 1, 3, 9
                    1, 5, 3, 4, 5, 9, 8 };
  
        // function call
        System.out.println(maximumValue(a, v));
    }
}
  
// This code is contributed by
// sanjeev2552


Python3




# Python3 code to find the maximum path sum 
dp = [0]*100
  
# Function for dfs traversal and
# to store the maximum value in
# dp[] for every node till the leaves 
def dfs(a, v, u, parent):
      
    # Initially dp[u] is always a[u]
    dp[u] = a[u - 1]
      
    # Stores the maximum value from nodes
    maximum = 0
      
    # Traverse the tree
    for child in v[u]:
          
        # If child is parent, then we continue 
        # without recursing further 
        if child == parent:
            continue
          
        # Call dfs for further traversal 
        dfs(a, v, child, u)
          
        # Store the maximum of previous visited  
        # node and present visited node 
        maximum = max(maximum, dp[child])
          
    # Add the maximum value 
    # returned to the parent node 
    dp[u] += maximum
  
  
# Function that returns the maximum value
def maximumValue(a, v):
    dfs(a, v, 1, 0)
    return dp[1]
  
# Driver Code 
def main():
      
    # Number of nodes 
    n = 14
      
    # Adjacency list as a dictionary
    v = {}
    for i in range(n + 1):
        v[i] = []
          
    # Create undirected edges 
    # initialize the tree given in the diagram 
    v[1].append(2), v[2].append(1)
    v[1].append(3), v[3].append(1
    v[1].append(4), v[4].append(1
    v[2].append(5), v[5].append(2
    v[2].append(6), v[6].append(2
    v[3].append(7), v[7].append(3
    v[4].append(8), v[8].append(4
    v[4].append(9), v[9].append(4
    v[4].append(10), v[10].append(4
    v[5].append(11), v[11].append(5
    v[5].append(12), v[12].append(5
    v[7].append(13), v[13].append(7
    v[7].append(14), v[14].append(7
      
    # Values of node 1, 2, 3....14 
    a = [ 3, 2, 1, 10, 1, 3, 9,
          1, 5, 3, 4, 5, 9, 8 ]
      
    # Function call
    print(maximumValue(a, v))
main()
  
# This code is contributed by stutipathak31jan  


C#




// C# code to find the maximum path sum 
using System;
using System.Collections.Generic;
  
class GFG
{
    static int[] dp = new int[100];
  
    // function for dfs traversal and to 
    // store the maximum value in []dp 
    // for every node till the leaves
    static void dfs(int[] a, List<int>[] v, 
                    int u, int parent) 
    {
  
        // initially dp[u] is always a[u]
        dp[u] = a[u - 1];
  
        // stores the maximum value from nodes
        int maximum = 0;
  
        // traverse the tree
        foreach(int child in v[u]) 
        {
  
            // if child is parent, then we continue
            // without recursing further
            if (child == parent)
                continue;
  
            // call dfs for further traversal
            dfs(a, v, child, u);
  
            // store the maximum of previous visited 
            // node and present visited node
            maximum = Math.Max(maximum, dp[child]);
        }
  
        // add the maximum value returned
        // to the parent node
        dp[u] += maximum;
    }
  
    // function that returns the maximum value
    static int maximumValue(int[] a, 
                            List<int>[] v)
    {
        dfs(a, v, 1, 0);
        return dp[1];
    }
  
    // Driver Code
    public static void Main(String[] args) 
    {
  
        // Driver Code
        int n = 14;
  
        // adjacency list
  
        List<int>[] v = new List<int>[n + 1];
  
        for (int i = 0; i < v.Length; i++)
            v[i] = new List<int>();
  
        // create undirected edges
        // initialize the tree given in the diagram
        v[1].Add(2); v[2].Add(1);
        v[1].Add(3); v[3].Add(1);
        v[1].Add(4); v[4].Add(1);
        v[2].Add(5); v[5].Add(2);
        v[2].Add(6); v[6].Add(2);
        v[3].Add(7); v[7].Add(3);
        v[4].Add(8); v[8].Add(4);
        v[4].Add(9); v[9].Add(4);
        v[4].Add(10); v[10].Add(4);
        v[5].Add(11); v[11].Add(5);
        v[5].Add(12); v[12].Add(5);
        v[7].Add(13); v[13].Add(7);
        v[7].Add(14); v[14].Add(7);
  
        // values of node 1, 2, 3....14
        int []a = { 3, 2, 1, 10, 1, 3, 9, 
                    1, 5, 3, 4, 5, 9, 8 };
  
        // function call
        Console.WriteLine(maximumValue(a, v));
    }
}
  
// This code is contributed by PrinciRaj1992


Javascript




<script>
  
// JavaScript code to find the maximum path sum 
var dp = Array(100).fill(0);
  
// function for dfs traversal and to 
// store the maximum value in []dp 
// for every node till the leaves
function dfs(a, v, u, parent) 
{
    // initially dp[u] is always a[u]
    dp[u] = a[u - 1];
    // stores the maximum value from nodes
    var maximum = 0;
    // traverse the tree
    for(var child of v[u]) 
    {
        // if child is parent, then we continue
        // without recursing further
        if (child == parent)
            continue;
        // call dfs for further traversal
        dfs(a, v, child, u);
        // store the maximum of previous visited 
        // node and present visited node
        maximum = Math.max(maximum, dp[child]);
    }
    // add the maximum value returned
    // to the parent node
    dp[u] += maximum;
}
// function that returns the maximum value
function maximumValue(a, v)
{
    dfs(a, v, 1, 0);
    return dp[1];
}
// Driver Code
// Driver Code
var n = 14;
// adjacency list
var v = Array.from(Array(n+1), ()=>Array());
for (var i = 0; i < v.length; i++)
    v[i] = [];
// create undirected edges
// initialize the tree given in the diagram
v[1].push(2); v[2].push(1);
v[1].push(3); v[3].push(1);
v[1].push(4); v[4].push(1);
v[2].push(5); v[5].push(2);
v[2].push(6); v[6].push(2);
v[3].push(7); v[7].push(3);
v[4].push(8); v[8].push(4);
v[4].push(9); v[9].push(4);
v[4].push(10); v[10].push(4);
v[5].push(11); v[11].push(5);
v[5].push(12); v[12].push(5);
v[7].push(13); v[13].push(7);
v[7].push(14); v[14].push(7);
// values of node 1, 2, 3....14
var a = [3, 2, 1, 10, 1, 3, 9, 
            1, 5, 3, 4, 5, 9, 8];
// function call
document.write(maximumValue(a, v));
  
</script>


Output

22

Time Complexity: O(N), where N is the number of nodes.
Auxiliary Space: O(N), for creating an additional dp array.



Last Updated : 22 Dec, 2022
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