Find the kth node in vertical order traversal of a Binary Tree

Given a binary tree and an integer k, the task is to print the kth node in the vertical order traversal of binary tree.If no such node exists then print -1.

The vertical order traversal of a binary tree means to print it vertically.

Examples:

Input: 
           1
         /   \
       2       3
     /  \     /  \
   4     5   6    7
              \    \
               8    9  
k = 3
Output: 1
The vertical order traversal of above tree is:
4
2
1 5 6
3 8
7
9

Input:
           1
         /   \
       2       3
     /  \     /  \
   4     5   6    7
              \    \
               8    9
k = 13
Output: -1


Approach: The idea is to perform vertical order traversal and check if the current node is the kth node then print its value, if number of nodes in the tree is less than K then print -1.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Structure for a binary tree node
struct Node {
    int key;
    Node *left, *right;
};
  
// A utility function to create a new node
Node* newNode(int key)
{
    Node* node = new Node;
    node->key = key;
    node->left = node->right = NULL;
    return node;
}
  
// Function to find kth node
// in vertcial order traversal
int KNodeVerticalOrder(Node* root, int k)
{
    // Base case
    if (!root || k == 0)
        return -1;
  
    int n = 0;
  
    // Variable to store kth node
    int k_node = -1;
  
    // Create a map and store vertical order in
    // map
    map<int, vector<int> > m;
    int hd = 0;
  
    // Create queue to do level order traversal
    // Every item of queue contains node and
    // horizontal distance
    queue<pair<Node*, int> > que;
    que.push(make_pair(root, hd));
  
    while (!que.empty()) {
        // Pop from queue front
        pair<Node*, int> temp = que.front();
        que.pop();
        hd = temp.second;
        Node* node = temp.first;
  
        // Insert this node's data in vector of hash
        m[hd].push_back(node->key);
  
        if (node->left != NULL)
            que.push(make_pair(node->left, hd - 1));
        if (node->right != NULL)
            que.push(make_pair(node->right, hd + 1));
    }
  
    // Traverse the map and find kth
    // node
    map<int, vector<int> >::iterator it;
    for (it = m.begin(); it != m.end(); it++) {
        for (int i = 0; i < it->second.size(); ++i) {
            n++;
            if (n == k)
                return (it->second[i]);
        }
    }
  
    if (k_node == -1)
        return -1;
}
  
// Driver code
int main()
{
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
    root->right->right->right = newNode(9);
    root->right->right->left = newNode(10);
    root->right->right->left->right = newNode(11);
    root->right->right->left->right->right = newNode(12);
  
    int k = 5;
    cout << KNodeVerticalOrder(root, k);
  
    return 0;
}

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Python3

# Python3 implementation of the approach

# Tree node structure
class Node:

def __init__(self, key):
self.key = key
self.left = None
self.right = None

# Function to find kth node
# in vertcial order traversal
def KNodeVerticalOrder(root, k):

# Base case
if not root or k == 0:
return -1

n = 0

# Variable to store kth node
k_node = -1

# Create a map and store
# vertical order in map
m = {}
hd = 0

# Create queue to do level order
# traversal Every item of queue contains
# node and horizontal distance
que = []
que.append((root, hd))

while len(que) > 0:

# Pop from queue front
temp = que.pop(0)
hd = temp[1]
node = temp[0]

# Insert this node’s data in vector of hash
if hd not in m: m[hd] = []
m[hd].append(node.key)

if node.left != None:
que.append((node.left, hd – 1))
if node.right != None:
que.append((node.right, hd + 1))

# Traverse the map and find kth node
for it in sorted(m):
for i in range(0, len(m[it])):
n += 1
if n == k:
return m[it][i]

if k_node == -1:
return -1

# Driver code
if __name__ == “__main__”:

root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
root.right.left.right = Node(8)
root.right.right.right = Node(9)
root.right.right.left = Node(10)
root.right.right.left.right = Node(11)
root.right.right.left.right.right = Node(12)

k = 5
print(KNodeVerticalOrder(root, k))

# This code is contributed by Rituraj Jain

Output:

6


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Improved By : rituraj_jain