Check whether BST contains Dead End or not

Given a Binary search Tree that contains positive integer values greater than 0. The task is to check whether the BST contains a dead end or not. Here Dead End means, we are not able to insert any element after that node.

Examples: 

Input :        8
             /   \ 
           5      9
         /   \
        2     7
       /
      1               
Output : Yes
Explanation : Node "1" is the dead End because
         after that we cant insert any element.       

Input :       8
            /   \ 
           7     10
         /      /   \
        2      9     13

Output : Yes
Explanation : We can't insert any element at 
              node 9.  

If we take a closer look at the problem, we can notice that we basically need to check if there is a leaf node with value x such that x+1 and x-1 exist in BST with the exception of x = 1. For x = 1, we can’t insert 0 as the problem statement says BST contains positive integers only.

To implement the above idea we first traverse the whole BST and store all nodes in a set. We also store all leaves in a separate hash to avoid re-traversal of BST. Finally, we check for every leaf node x, if x-1 and x+1 are present in set or not.

Below is a C++ implementation of the above idea. 

C++

// C++ program check whether BST contains
// dead end or not
#include<bits/stdc++.h>
using namespace std;

// A BST node
struct Node
{
    int data;
    struct Node *left, *right;
};

// A utility function to create a new node
Node *newNode(int data)
{
    Node *temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

/* A utility function to insert a new Node
  with given key in BST */
struct Node* insert(struct Node* node, int key)
{
    /* If the tree is empty, return a new Node */
    if (node == NULL) return newNode(key);

    /* Otherwise, recur down the tree */
    if (key < node->data)
        node->left = insert(node->left, key);
    else if (key > node->data)
        node->right = insert(node->right, key);

    /* return the (unchanged) Node pointer */
    return node;
}

// Function to store all node of given binary search tree
void storeNodes(Node * root, unordered_set<int> &all_nodes,
                          unordered_set<int> &leaf_nodes)
{
    if (root == NULL)
        return ;

    // store all node of binary search tree
    all_nodes.insert(root->data);

    // store leaf node in leaf_hash
    if (root->left==NULL && root->right==NULL)
    {
        leaf_nodes.insert(root->data);
        return ;
    }

    // recur call rest tree
    storeNodes(root-> left, all_nodes, leaf_nodes);
    storeNodes(root->right, all_nodes, leaf_nodes);
}

// Returns true if there is a dead end in tree,
// else false.
bool isDeadEnd(Node *root)
{
    // Base case
    if (root == NULL)
        return false ;

    // create two empty hash sets that store all
    // BST elements and leaf nodes respectively.
    unordered_set<int> all_nodes, leaf_nodes;

    // insert 0 in 'all_nodes' for handle case
    // if bst contain value 1
    all_nodes.insert(0);

    // Call storeNodes function to store all BST Node
    storeNodes(root, all_nodes, leaf_nodes);

    // Traversal leaf node and check Tree contain
    // continuous sequence of
    // size tree or Not
    for (auto i = leaf_nodes.begin() ; i != leaf_nodes.end(); i++)
    {
        int x = (*i);

        // Here we check first and last element of
        // continuous sequence that are x-1 & x+1
        if (all_nodes.find(x+1) != all_nodes.end() &&
            all_nodes.find(x-1) != all_nodes.end())
            return true;
    }

    return false ;
}

// Driver program
int main()
{
/*       8
       /   \
      5    11
     /  \
    2    7
     \
      3
       \
        4 */
    Node *root = NULL;
    root = insert(root, 8);
    root = insert(root, 5);
    root = insert(root, 2);
    root = insert(root, 3);
    root = insert(root, 7);
    root = insert(root, 11);
    root = insert(root, 4);
    if (isDeadEnd(root) == true)
        cout << "Yes " << endl;
    else
        cout << "No " << endl;
    return 0;
}

Java

// Java program check whether BST contains
// dead end or not
import java.util.*;

class Main {
    // create two empty hash sets that store all
    // BST elements and leaf nodes respectively.
    static HashSet<Integer> all_nodes
        = new HashSet<Integer>();
    static HashSet<Integer> leaf_nodes
        = new HashSet<Integer>();

    /* A utility function to insert a new Node
    with given key in BST */
    public static Node insert(Node node, int key)
    {
        /* If the tree is empty, return a new Node */
        if (node == null)
            return new Node(key);

        /* Otherwise, recur down the tree */
        if (key < node.data)
            node.left = insert(node.left, key);
        else if (key > node.data)
            node.right = insert(node.right, key);

        /* return the Node */
        return node;
    }
    // Function to store all node of given binary search
    // tree
    public static void storeNodes(Node root)
    {
        if (root == null)
            return;

        // store all node of binary search tree
        all_nodes.add(root.data);

        // store leaf node in leaf_hash
        if (root.left == null && root.right == null) {
            leaf_nodes.add(root.data);
            return;
        }

        // recur call rest tree
        storeNodes(root.left);
        storeNodes(root.right);
    }

    // Returns true if there is a dead end in tree,
    // else false.
    public static boolean isDeadEnd(Node root)
    {
        // Base case
        if (root == null)
            return false;

        // insert 0 in 'all_nodes' for handle case
        // if bst contain value 1
        all_nodes.add(0);

        // Call storeNodes function to store all BST Node
        storeNodes(root);

        // Traversal leaf node and check Tree contain
        // continuous sequence of
        // size tree or Not
        for (int i : leaf_nodes) {
            int x = i;

            // Here we check first and last element of
            // continuous sequence that are x-1 & x+1
            if (all_nodes.contains(x + 1)
                && all_nodes.contains(x - 1)) {
                return true;
            }
        }
        return false;
    }

    // Driver program
    public static void main(String[] args)
    {
        /*       8
               /   \
              5    11
             /  \
            2    7
             \
              3
               \
                4 */
        Node root = null;
        root = insert(root, 8);
        root = insert(root, 5);
        root = insert(root, 2);
        root = insert(root, 3);
        root = insert(root, 7);
        root = insert(root, 11);
        root = insert(root, 4);
        if (isDeadEnd(root) == true)
            System.out.println("Yes");

        else
            System.out.println("No");
    }
}

// A BST node
class Node {
    int data;
    Node left, right;

    Node(int data)
    {
        this.data = data;
        left = null;
        right = null;
    }
}

// This code is contributed by Tapesh(tapeshdua420)

Python3

# Python 3 program check
# whether BST contains
# dead end or not
all_nodes = set()
leaf_nodes = set()

# A BST node
class newNode:

    def __init__(self, data):

        self.data = data
        self.left = None
        self.right = None


''' A utility function to
    insert a new Node with
    given key in BST '''
def insert(node, key):
    '''/* If the tree is empty,
          return a new Node */ '''
    if (node == None):
        return newNode(key)

    # Otherwise, recur down
    # the tree
    if (key < node.data):
        node.left = insert(node.left,
                           key)
    elif (key > node.data):
        node.right = insert(node.right,
                            key)

    # return the (unchanged)
    # Node pointer
    return node

# Function to store all node
# of given binary search tree
def storeNodes(root):

    global all_nodes
    global leaf_nodes
    if (root == None):
        return

    # store all node of binary
    # search tree
    all_nodes.add(root.data)

    # store leaf node in leaf_hash
    if (root.left == None and
            root.right == None):
        leaf_nodes.add(root.data)
        return

    # recur call rest tree
    storeNodes(root. left)
    storeNodes(root.right)

# Returns true if there is
# a dead end in tree,
# else false.
def isDeadEnd(root):

    global all_nodes
    global leaf_nodes

    # Base case
    if (root == None):
        return False

    # create two empty hash
    # sets that store all BST
    # elements and leaf nodes
    # respectively.

    # insert 0 in 'all_nodes'
    # for handle case if bst
    # contain value 1
    all_nodes.add(0)

    # Call storeNodes function
    # to store all BST Node
    storeNodes(root)

    # Traversal leaf node and
    # check Tree contain
    # continuous sequence of
    # size tree or Not
    for x in leaf_nodes:

        # Here we check first and
        # last element of continuous
        # sequence that are x-1 & x+1
        if ((x + 1) in all_nodes and
                (x - 1) in all_nodes):
            return True

    return False


# Driver code
if __name__ == '__main__':

    '''/*       8
       /   \
      5    11
     /  \
    2    7
     \
      3
       \
        4 */
    '''
    root = None
    root = insert(root, 8)
    root = insert(root, 5)
    root = insert(root, 2)
    root = insert(root, 3)
    root = insert(root, 7)
    root = insert(root, 11)
    root = insert(root, 4)

    if(isDeadEnd(root) == True):
        print("Yes")
    else:
        print("No")

# This code is contributed by bgangwar59

Output

Yes 

Time Complexity : O(n)

Improved Approach

In the above approach we are using 2 hashmaps , one for storing all nodes and one for storing leaf nodes , instead of using 2 maps we can do this problem with 1 hashmap also .

We can pass the hashmap of all nodes and check if for node x there exists a x-1 and x+1 or not. If we got a node for which x+1 and x-1 both are present we will return true otherwise we will return false

Implementation:

C++

// C++ program check whether BST contains
// dead end or not
#include<bits/stdc++.h>
using namespace std;

// A BST node
struct Node
{
    int data;
    struct Node *left, *right;
};

// A utility function to create a new node
Node *newNode(int data)
{
    Node *temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

/* A utility function to insert a new Node
  with given key in BST */
struct Node* insert(struct Node* node, int key)
{
    /* If the tree is empty, return a new Node */
    if (node == NULL) return newNode(key);

    /* Otherwise, recur down the tree */
    if (key < node->data)
        node->left = insert(node->left, key);
    else if (key > node->data)
        node->right = insert(node->right, key);

    /* return the (unchanged) Node pointer */
    return node;
}
void findallNodes(Node* root,map<int,int> &allnodes)
{
    if(root == NULL)
    return ;

    allnodes[root->data] = 1;
    findallNodes(root->left,allnodes);
    findallNodes(root->right,allnodes);
}
bool check(Node* root,map<int,int> &allnodes)
{
    if(root == NULL)
    return false;

    if(root->left == NULL and root->right == NULL)
    {
        int pre = root->data - 1;
        int next = root->data + 1;

        if(allnodes.find(pre) != allnodes.end() and allnodes.find(next) != allnodes.end())
        return true;
    }

    return check(root->left,allnodes) or check(root->right,allnodes);

}
bool isDeadEnd(Node *root)
{
    // Base case
   if (root == NULL)
        return false ;
    map<int,int> allnodes;
      // adding 0 for handling the exception of node having data = 1
    allnodes[0] = 1;
    findallNodes(root,allnodes);

    return check(root,allnodes);

}

// Driver program
int main()
{
/*       8
       /   \
      5    11
     /  \
    2    7
     \
      3
       \
        4 */
    Node *root = NULL;
    root = insert(root, 8);
    root = insert(root, 5);
    root = insert(root, 2);
    root = insert(root, 3);
    root = insert(root, 7);
    root = insert(root, 11);
    root = insert(root, 4);
    if (isDeadEnd(root) == true)
        cout << "Yes " << endl;
    else
        cout << "No " << endl;
    return 0;
}

Output

Yes 

Simple Recursive solution to check whether BST contains dead End
This article is contributed by Nishant_Singh(Pintu). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.