Print the nodes that are just above the leaf node

Given a binary tree consisting of N nodes, the task is to print the nodes that are just above the leaf node.

Examples:

Input: N = 7, Below is the given Binary Tree:

Output: 20 8 12
Explanation:
Node 20 is just above the leaf node 22.
Node 8 is just above the leaf node 4.
Node 12 is just above the leaf nodes 10 and 14.

Input: N = 5, Below is the given Binary Tree:

Output: 1 2
Explanation:
Node 1 is just above the leaf node 3.
Node 2 is just above the leaf nodes 4 and 5.

Approach: The idea is to traverse the tree and for each node, check if it can be the one which is just above the leaf node. For that, the current node must have children and at least one of the children should be a leaf node. Below are the steps:



  • Traverse the tree and check each node.
  • If the current node has two children, check if any of them is a root node. If yes, then print the current node.
  • If the current node has only left or right child, then check if that left or the right child is a leaf node. If yes, then print the current node.
  • Else, continue traversing the tree and move to next node.

Below is the implementation of the above approach:

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Node of tree
struct node {
    int data;
    struct node *left, *right;
};
  
// Creates and initializes a new
// node for the tree
struct node* newnode(int data)
{
    struct node* temp = new node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
  
// Prints all nodes which are just
// above leaf node
void cal(struct node* root)
{
  
    // If tree is empty
    if (root == NULL) {
        return;
    }
  
    // If it is a leaf node
    if (root->left == NULL
        && root->right == NULL) {
        return;
    }
  
    // For internal nodes
    else {
  
        // If node has two children
        if (root->left != NULL
            && root->right != NULL) {
  
            if ((root->left->left == NULL
                 && root->left->right == NULL)
                || (root->right->left == NULL
                    && root->right->right == NULL)) {
  
                cout << root->data << " ";
            }
        }
  
        // If node has only left child
        if (root->left != NULL
            && root->right == NULL) {
  
            if (root->left->left == NULL
                && root->left->right == NULL) {
  
                cout << root->data << " ";
            }
        }
  
        // If node has only right child
        if (root->right != NULL
            && root->left == NULL) {
  
            if (root->right->left == NULL
                && root->right->right == NULL) {
  
                cout << root->data << " ";
            }
        }
    }
  
    // Recursively Call for left
    // and right subtree
    cal(root->left);
    cal(root->right);
}
  
// Driver Code
int main()
{
    // Construct a tree
    node* root = newnode(20);
    root->left = newnode(8);
    root->right = newnode(22);
    root->left->left = newnode(4);
    root->left->right = newnode(12);
    root->left->right->left = newnode(10);
    root->left->right->right = newnode(14);
  
    // Function Call
    cal(root);
  
    return 0;
}

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Output:

20 8 12

Time Complexity: O(N)
Auxiliary Space: O(1)

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