Print the nodes of binary tree as they become the leaf node

Given a binary tree. First print all leaf nodes, after that remove all the leaf nodes from the tree and now print all the new formed leaf nodes and keep doing this until all the nodes are removed from the tree.

Examples :

Input :  8
             / \
           3    10
          / \     / \
         1  6  14 4
        / \
       7   13

Output : 
4 6 7 13 14
1 10
3
8


Source :Flipkart On Campus Recruitment

Approach : The idea is to perform simple dfs and assign different values to each node on the basis of following conditions:

  1. Initially assign all the nodes with value as 0.
  2. Now, Assign all the nodes with the value as (maximum value of both child)+1.

Tree before DFS: A temporary value zero is assigned to all of the nodes.
before dfs

Tree after DFS: Nodes are assigned with the value as (maximum value of both child)+1.
after dfs

Now, you can see in the above tree that after all the values are assigned to each node, the task now reduces to print the tree on the basis of increasing order of node values assigned to them.

Below is the implementation of above approach:

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// C++ program to print the nodes of binary
// tree as they become the leaf node
  
#include <bits/stdc++.h>
using namespace std;
  
// Binary tree node
struct Node {
    int data;
    int order;
    struct Node* left;
    struct Node* right;
};
  
// Utiltiy function to allocate a new node
struct Node* newNode(int data, int order)
{
    struct Node* node = new Node;
    node->data = data;
    node->order = order;
    node->left = NULL;
    node->right = NULL;
  
    return (node);
}
  
// Function for inorder traversal of tree and
// assigning values to nodes
void Inorder(struct Node* node, vector<pair<int, int> >& v)
{
    if (node == NULL)
        return;
  
    /* first recur on left child */
    Inorder(node->left, v);
  
    /* now recur on right child */
    Inorder(node->right, v);
  
    // If current node is leaf node, it's order will be 1
    if (node->right == NULL && node->left == NULL) {
        node->order = 1;
  
        // make pair of assigned value and tree value
        v.push_back(make_pair(node->order, node->data));
    }
    else {
        // otherwise, the order will be:
        // max(left_child_order, right_child_order) + 1
        node->order = max((node->left)->order, (node->right)->order) + 1;
  
        // make pair of assigned value and tree value
        v.push_back(make_pair(node->order, node->data));
    }
}
  
// Function to print leaf nodes in
// the given order
void printLeafNodes(int n, vector<pair<int, int> >& v)
{
    // Sort the vector in increasing order of
    // assigned node values
    sort(v.begin(), v.end());
  
    for (int i = 0; i < n; i++) {
        if (v[i].first == v[i + 1].first)
            cout << v[i].second << " ";
  
        else
            cout << v[i].second << "\n";
    }
}
  
// Driver Code
int main()
{
    struct Node* root = newNode(8, 0);
    root->left = newNode(3, 0);
    root->right = newNode(10, 0);
    root->left->left = newNode(1, 0);
    root->left->right = newNode(6, 0);
    root->right->left = newNode(14, 0);
    root->right->right = newNode(4, 0);
    root->left->left->left = newNode(7, 0);
    root->left->left->right = newNode(13, 0);
  
    int n = 9;
  
    vector<pair<int, int> > v;
  
    Inorder(root, v);
    printLeafNodes(n, v);
  
    return 0;
}

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Output:

4 6 7 13 14
1 10
3
8

Time Complexity : O(nlogn)
Auxiliary Space : O(n), where n is the number of nodes in the given Binary Tree.



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