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Print the nodes that are just above the leaf node
• Last Updated : 11 Nov, 2020

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 a 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 the next node.

Below is the implementation of the above approach:

## C++14

 // C++ program for the above approach #include 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; }

## Java

 // Java program for the above approach import java.util.*;   // Class containing the left and right // child of current node and the // key value class Node {     int data;     Node left, right;       // Constructor of the class     public Node(int item)     {         data = item;         left = right = null;     } }   class GFG{       Node root;   // Prints all nodes which are just // above leaf node static void cal(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))             {                 System.out.print(root.data + " ");             }         }           // If node has only left child         if (root.left != null &&            root.right == null)         {             if (root.left.left == null &&                root.left.right == null)             {                 System.out.print(root.data + " ");             }         }           // If node has only right child         if (root.right != null &&              root.left == null)         {             if (root.right.left == null &&                root.right.right == null)             {                 System.out.print(root.data + " ");             }         }     }       // Recursively call for left     // and right subtree     cal(root.left);     cal(root.right); }       // Driver Code public static void main (String[] args) {     GFG tree = new GFG();       tree.root = new Node(20);     tree.root.left = new Node(8);     tree.root.right = new Node(22);       tree.root.left.left = new Node(4);     tree.root.left.right = new Node(12);       tree.root.left.right.left = new Node(10);     tree.root.left.right.right = new Node(14);           // Function call     cal(tree.root); } }   // This code is contributed by offbeat

## Python3

 # Python3 program for the # above approach   # Node of tree class newNode:         def __init__(self, data):                 self.data = data         self.left = None         self.right = None   # Creates and initializes a new # node for the tree   # Prints all nodes which are # just above leaf node def cal(root):         # If tree is empty     if (root == None):         return       # If it is a leaf node     if (root.left == None and         root.right == None):         return       # For internal nodes     else:                 # If node has two children         if (root.left != None and             root.right != None):             if ((root.left.left == None and                  root.left.right == None) or                 (root.right.left == None and                  root.right.right == None)):                 print(root.data, end = " ")           # If node has only left child         if (root.left != None and             root.right == None):             if (root.left.left == None and                 root.left.right == None):                 print(root.data, end = " ")           # If node has only right child         if (root.right != None and             root.left == None):             if (root.right.left == None and                 root.right.right == None):                 print(root.data, end = " ")       # Recursively Call for left     # and right subtree     cal(root.left)     cal(root.right)   # Driver Code if __name__ == '__main__':         # Construct a tree     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)   # This code is contributed by SURENDRA_GANGWAR

## C#

 // C# program for the above approach using System;   // Class containing the left and right // child of current node and the // key value class Node {     public int data;     public Node left, right;       // Constructor of the class     public Node(int item)     {         data = item;         left = right = null;     } }   class GFG{    Node root;   // Prints all nodes which are just // above leaf node static void cal(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))             {                 Console.Write(root.data + " ");             }         }           // If node has only left child         if (root.left != null &&             root.right == null)         {             if (root.left.left == null &&                 root.left.right == null)             {                 Console.Write(root.data + " ");             }         }           // If node has only right child         if (root.right != null &&             root.left == null)         {             if (root.right.left == null &&                 root.right.right == null)             {                 Console.Write(root.data + " ");             }         }     }       // Recursively call for left     // and right subtree     cal(root.left);     cal(root.right); }       // Driver Code public static void Main(String[] args) {     GFG tree = new GFG();     tree.root = new Node(20);     tree.root.left = new Node(8);     tree.root.right = new Node(22);     tree.root.left.left = new Node(4);     tree.root.left.right = new Node(12);     tree.root.left.right.left = new Node(10);     tree.root.left.right.right = new Node(14);           // Function call     cal(tree.root); } }   // This code is contributed by Rajput-Ji

Output:

20 8 12

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

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