Print all leaf nodes of an n-ary tree using DFS

Given an array edge[][2] where (edge[i][0], edge[i][1]) defines an edge in the n-ary tree, the task is to print all the leaf nodes of the given tree using.

Examples:

Input: edge[][] = {{1, 2}, {1, 3}, {2, 4}, {2, 5}, {3, 6}}
Output: 4 5 6
    1
   / \
  2   3
 / \   \
4   5   6

Input: edge[][] = {{1, 5}, {1, 7}, {5, 6}}
Output: 6 7

Approach: DFS can be used to traverse the complete tree. We will keep track of parent while traversing to avoid the visited node array. Initially for every node we can set a flag and if the node have at least one child (i.e. non-leaf node) then we will reset the flag. The nodes with no children are the leaf nodes.



Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to perform DFS on the tree
void dfs(list<int> t[], int node, int parent)
{
    int flag = 1;
  
    // Iterating the children of current node
    for (auto ir : t[node]) {
  
        // There is at least a child
        // of the current node
        if (ir != parent) {
            flag = 0;
            dfs(t, ir, node);
        }
    }
  
    // Current node is connected to only
    // its parent i.e. it is a leaf node
    if (flag == 1)
        cout << node << " ";
}
  
// Driver code
int main()
{
    // Adjacency list
    list<int> t[1005];
  
    // List of all edges
    pair<int, int> edges[] = { { 1, 2 },
                               { 1, 3 },
                               { 2, 4 },
                               { 3, 5 },
                               { 3, 6 },
                               { 3, 7 },
                               { 6, 8 } };
  
    // Count of edges
    int cnt = sizeof(edges) / sizeof(edges[0]);
  
    // Number of nodes
    int node = cnt + 1;
  
    // Create the tree
    for (int i = 0; i < cnt; i++) {
        t[edges[i].first].push_back(edges[i].second);
        t[edges[i].second].push_back(edges[i].first);
    }
  
    // Function call
    dfs(t, 1, 0);
  
    return 0;
}

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Python3

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# Python3 implementation of the approach
t = [[] for i in range(1005)]
  
# Function to perform DFS on the tree
def dfs(node, parent):
    flag = 1
  
    # Iterating the children of current node
    for ir in t[node]:
  
        # There is at least a child
        # of the current node
        if (ir != parent):
            flag = 0
            dfs(ir, node)
  
    # Current node is connected to only
    # its parent i.e. it is a leaf node
    if (flag == 1):
        print(node, end = " ")
  
# Driver code
  
# List of all edges
edges = [[ 1, 2 ],
         [ 1, 3 ],
         [ 2, 4 ],
         [ 3, 5 ],
         [ 3, 6 ],
         [ 3, 7 ],
         [ 6, 8 ]]
  
# Count of edges
cnt = len(edges)
  
# Number of nodes
node = cnt + 1
  
# Create the tree
for i in range(cnt):
    t[edges[i][0]].append(edges[i][1])
    t[edges[i][1]].append(edges[i][0])
  
# Function call
dfs(1, 0)
  
# This code is contributed by Mohit Kumar

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Output:

4 5 8 7


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Improved By : mohit kumar 29



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