Print Nodes of Binary Tree as They Become Leaf Node in C++ Programming

Given a binary tree, we have to print its leaf nodes and then we have to remove those leaf nodes and then repeat till there are no nodes left in the tree.

Example

So the output of the problem should be −

6 7 9 13 14
3 4
2
1

Approach

We have adopted an approach where we are applying DFS.

For applying a temporary value zero is assigned to every value and then assign all the nodes with the value maximum(value of both child)+1.

Algorithm

START
STEP 1-> DEFINE A struct Node
   WITH DATA MEMBERS data, order, *left, *right
STEP 2-> DEFINE A struct Node* newNode(int data, int order)
   WITH struct Node* node = new Node, node->data = data, node->order = order,
   node->left = NULL, node->right = NULL, return (node)
FUNCTION void postod(struct Node* node, vector<pair<int, int> >& v)
STEP 1-> IF node == NULL THEN,
   RETURN
STEP 2-> CALL FUNCTION postod(node->left, v)
STEP 3-> CALL FUNCTION postod(node->right, v)
STEP 4-> IF node->right == NULL && node->left == NULL THEN,
   SET node->order AS 1
   v.push_back(make_pair(node->order, node->data))
   ELSE
      node->order = max((node->left)->order, (node->right)->order) + 1
   v.push_back(make_pair(node->order, node->data))
END IF
FUNCTION void printLeafNodes(int n, vector<pair<int, int> >& v)
STEP 1-> sort(v.begin(), v.end())
STEP 2-> LOOP FOR i = 0 AND i < n AND i++
   IF v[i].first == v[i + 1].first THEN,
      PRINT v[i].second
   ELSE
      PRINT v[i].second
   END IF
END FOR
IN main()
STEP 1-> CREATE A ROOT NODE LIKE struct Node* root = newNode(8, 0)
STEP 2-> DECLARE AND SET n = 9
STEP 3-> CALL postod(root, v);
STEP 4-> CALL printLeafNodes(n, v);

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
struct Node {
   int data;
   int order;
   struct Node* left;
   struct Node* right;
};
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);
}
void postod(struct Node* node, vector<pair<int, int> >& v){
   if (node == NULL)
      return;
      /* first recur on left child */
   postod(node->left, v);
   /* now recur on right child */
   postod(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 {
      node->order = max((node->left)->order, (node->right)->order) + 1;
      v.push_back(make_pair(node->order, node->data));
   }
}
void printLeafNodes(int n, vector<pair<int, int> >& v){
   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";
   }
}
int main(){
   struct Node* root = newNode(1, 0);
   root->left = newNode(2, 0);
   root->right = newNode(3, 0);
   root->left->left = newNode(4, 0);
   root->left->right = newNode(6, 0);
   root->right->left = newNode(14, 0);
   root->right->right = newNode(9, 0);
   root->left->left->left = newNode(7, 0);
   root->left->left->right = newNode(13, 0);
   int n = 9;
   vector<pair<int, int> > v;
   postod(root, v);
   printLeafNodes(n, v);
   return 0;
}

Output

This Program will Print output −

6 7 9 13 14
3 4
2
1