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Find the closest leaf in a Binary Tree

  • Difficulty Level : Hard
  • Last Updated : 11 Nov, 2021

Given a Binary Tree and a key ‘k’, find distance of the closest leaf from ‘k’.

Examples: 

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              A
            /    \    
           B       C
                 /   \  
                E     F   
               /       \
              G         H
             / \       /
            I   J     K

Closest leaf to 'H' is 'K', so distance is 1 for 'H'
Closest leaf to 'C' is 'B', so distance is 2 for 'C'
Closest leaf to 'E' is either 'I' or 'J', so distance is 2 for 'E' 
Closest leaf to 'B' is 'B' itself, so distance is 0 for 'B' 

The main point to note here is that a closest key can either be a descendant of given key or can be reached through one of the ancestors. 



The idea is to traverse the given tree in preorder and keep track of ancestors in an array. When we reach the given key, we evaluate distance of the closest leaf in subtree rooted with given key. We also traverse all ancestors one by one and find distance of the closest leaf in the subtree rooted with ancestor. We compare all distances and return minimum.

Below is the implementation of above approach.  

C++




// A C++ program to find the closest leaf of a given key in Binary Tree
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree Node has key, pocharer to left and right children */
struct Node
{
    char key;
    struct Node* left, *right;
};
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pocharers. */
Node *newNode(char k)
{
    Node *node = new Node;
    node->key = k;
    node->right = node->left = NULL;
    return node;
}
 
// A utility function to find minimum of x and y
int getMin(int x, int y)
{
    return (x < y)? x :y;
}
 
// A utility function to find distance of closest leaf of the tree
// rooted under given root
int closestDown(struct Node *root)
{
    // Base cases
    if (root == NULL)
        return INT_MAX;
    if (root->left == NULL && root->right == NULL)
        return 0;
 
    // Return minimum of left and right, plus one
    return 1 + getMin(closestDown(root->left), closestDown(root->right));
}
 
// Returns distance of the closest leaf to a given key 'k'.  The array
// ancestors is used to keep track of ancestors of current node and
// 'index' is used to keep track of current index in 'ancestors[]'
int findClosestUtil(struct Node *root, char k, struct Node *ancestors[],
                                               int index)
{
    // Base case
    if (root == NULL)
        return INT_MAX;
 
    // If key found
    if (root->key == k)
    {
        //  Find the closest leaf under the subtree rooted with given key
        int res = closestDown(root);
 
        // Traverse all ancestors and update result if any parent node
        // gives smaller distance
        for (int i = index-1; i>=0; i--)
            res = getMin(res, index - i + closestDown(ancestors[i]));
        return res;
    }
 
    // If key node found, store current node and recur for left and
    // right childrens
    ancestors[index] = root;
    return getMin(findClosestUtil(root->left, k, ancestors, index+1),
                  findClosestUtil(root->right, k, ancestors, index+1));
 
}
 
// The main function that returns distance of the closest key to 'k'. It
// mainly uses recursive function findClosestUtil() to find the closes
// distance.
int findClosest(struct Node *root, char k)
{
    // Create an array to store ancestors
    // Assumption: Maximum height of tree is 100
    struct Node *ancestors[100];
 
    return findClosestUtil(root, k, ancestors, 0);
}
 
/* Driver program to test above functions*/
int main()
{
    // Let us construct the BST shown in the above figure
    struct Node *root        = newNode('A');
    root->left               = newNode('B');
    root->right              = newNode('C');
    root->right->left        = newNode('E');
    root->right->right       = newNode('F');
    root->right->left->left  = newNode('G');
    root->right->left->left->left  = newNode('I');
    root->right->left->left->right = newNode('J');
    root->right->right->right      = newNode('H');
    root->right->right->right->left = newNode('K');
 
    char k = 'H';
    cout << "Distance of the closest key from " << k << " is "
         << findClosest(root, k) << endl;
    k = 'C';
    cout << "Distance of the closest key from " << k << " is "
         << findClosest(root, k) << endl;
    k = 'E';
    cout << "Distance of the closest key from " << k << " is "
         << findClosest(root, k) << endl;
    k = 'B';
    cout << "Distance of the closest key from " << k << " is "
         << findClosest(root, k) << endl;
 
    return 0;
}

Java




// Java program to find closest leaf of a given key in Binary Tree
  
/* Class containing left and right child of current
   node and key value*/
class Node
{
    int data;
    Node left, right;
  
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}
  
class BinaryTree
{
    Node root;
      
    // A utility function to find minimum of x and y
    int getMin(int x, int y)
    {
        return (x < y) ? x : y;
    }
  
    // A utility function to find distance of closest leaf of the tree
    // rooted under given root
    int closestDown(Node node)
    {
        // Base cases
        if (node == null)
            return Integer.MAX_VALUE;
        if (node.left == null && node.right == null)
            return 0;
  
        // Return minimum of left and right, plus one
        return 1 + getMin(closestDown(node.left), closestDown(node.right));
    }
  
    // Returns distance of the closest leaf to a given key 'k'.  The array
    // ancestors is used to keep track of ancestors of current node and
    // 'index' is used to keep track of current index in 'ancestors[]'
    int findClosestUtil(Node node, char k, Node ancestors[], int index)
    {
        // Base case
        if (node == null)
            return Integer.MAX_VALUE;
  
        // If key found
        if (node.data == k)
        {
            //  Find the closest leaf under the subtree rooted with given key
            int res = closestDown(node);
  
            // Traverse all ancestors and update result if any parent node
            // gives smaller distance
            for (int i = index - 1; i >= 0; i--)
                res = getMin(res, index - i + closestDown(ancestors[i]));
            return res;
        }
  
        // If key node found, store current node and recur for left and
        // right childrens
        ancestors[index] = node;
        return getMin(findClosestUtil(node.left, k, ancestors, index + 1),
                findClosestUtil(node.right, k, ancestors, index + 1));
  
    }
  
    // The main function that returns distance of the closest key to 'k'. It
    // mainly uses recursive function findClosestUtil() to find the closes
    // distance.
    int findClosest(Node node, char k)
    {
        // Create an array to store ancestors
        // Assumption: Maximum height of tree is 100
        Node ancestors[] = new Node[100];
  
        return findClosestUtil(node, k, ancestors, 0);
    }
  
    // Driver program to test for above functions
    public static void main(String args[])
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node('A');
        tree.root.left = new Node('B');
        tree.root.right = new Node('C');
        tree.root.right.left = new Node('E');
        tree.root.right.right = new Node('F');
        tree.root.right.left.left = new Node('G');
        tree.root.right.left.left.left = new Node('I');
        tree.root.right.left.left.right = new Node('J');
        tree.root.right.right.right = new Node('H');
        tree.root.right.right.right.left = new Node('H');
  
        char k = 'H';
        System.out.println("Distance of the closest key from " + k + " is "
                            + tree.findClosest(tree.root, k));
        k = 'C';
        System.out.println("Distance of the closest key from " + k + " is "
                            + tree.findClosest(tree.root, k));
        k = 'E';
        System.out.println("Distance of the closest key from " + k + " is "
                            + tree.findClosest(tree.root, k));
        k = 'B';
        System.out.println("Distance of the closest key from " + k + " is "
                             + tree.findClosest(tree.root, k));
  
    }
}
  
// This code has been contributed by Mayank Jaiswal

Python




# Python program to find closest leaf of a
# given key in binary tree
 
INT_MAX = 2**32
 
# A binary tree node
class Node:
    # Constructor to create a binary tree
    def __init__(self ,key):
        self.key = key
        self.left  = None
        self.right = None
 
def closestDown(root):
    #Base Case
    if root is None:
        return INT_MAX
    if root.left is None and root.right is None:
        return 0
     
    # Return minimum of left and right plus one
    return 1 + min(closestDown(root.left),
                   closestDown(root.right))
 
# Returns distance of the closes leaf to a given key k
# The array ancestors us used to keep track of ancestors
# of current node and 'index' is used to keep track of
# current index in 'ancestors[i]'
def findClosestUtil(root, k, ancestors, index):
    # Base Case
    if root is None:
        return INT_MAX
     
    # if key found
    if root.key == k:
        # Find closest leaf under the subtree rooted
        # with given key
        res = closestDown(root)
         
        # Traverse ll ancestors and update result if any
        # parent node gives smaller distance
        for i in reversed(range(0,index)):
            res = min(res, index-i+closestDown(ancestors[i]))
        return res
 
    # if key node found, store current node and recur for left
    # and right childrens
    ancestors[index] = root
    return min(
        findClosestUtil(root.left, k,ancestors, index+1),
        findClosestUtil(root.right, k, ancestors, index+1))
 
# The main function that return distance of the clses key to
# 'key'. It mainly uses recursive function findClosestUtil()
# to find the closes distance
def findClosest(root, k):
    # Create an array to store ancestors
    # Assumption: Maximum height of tree is 100
    ancestors = [None for i in range(100)]
 
    return findClosestUtil(root, k, ancestors, 0)
 
 
# Driver program to test above function
root = Node('A')
root.left = Node('B')
root.right = Node('C');
root.right.left = Node('E');
root.right.right  = Node('F');
root.right.left.left = Node('G');
root.right.left.left.left  = Node('I');
root.right.left.left.right = Node('J');
root.right.right.right  = Node('H');
root.right.right.right.left = Node('K');
 
k = 'H';
print "Distance of the closest key from "+ k + " is",
print findClosest(root, k)
 
k = 'C'
print "Distance of the closest key from " + k + " is",
print findClosest(root, k)
 
k = 'E'
print "Distance of the closest key from " + k + " is",
print findClosest(root, k)
 
k = 'B'
print "Distance of the closest key from " + k + " is",
print findClosest(root, k)
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#




using System;
 
// C# program to find closest leaf of a given key in Binary Tree
 
/* Class containing left and right child of current 
   node and key value*/
public class Node
{
    public int data;
    public Node left, right;
 
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}
 
public class BinaryTree
{
    public Node root;
 
    // A utility function to find minimum of x and y
    public virtual int getMin(int x, int y)
    {
        return (x < y) ? x : y;
    }
 
    // A utility function to find distance of closest leaf of the tree
    // rooted under given root
    public virtual int closestDown(Node node)
    {
        // Base cases
        if (node == null)
        {
            return int.MaxValue;
        }
        if (node.left == null && node.right == null)
        {
            return 0;
        }
 
        // Return minimum of left and right, plus one
        return 1 + getMin(closestDown(node.left), closestDown(node.right));
    }
 
    // Returns distance of the closest leaf to a given key 'k'.  The array
    // ancestors is used to keep track of ancestors of current node and
    // 'index' is used to keep track of current index in 'ancestors[]'
    public virtual int findClosestUtil(Node node, char k, Node[] ancestors, int index)
    {
        // Base case
        if (node == null)
        {
            return int.MaxValue;
        }
 
        // If key found
        if ((char)node.data == k)
        {
            //  Find the closest leaf under the subtree rooted with given key
            int res = closestDown(node);
 
            // Traverse all ancestors and update result if any parent node
            // gives smaller distance
            for (int i = index - 1; i >= 0; i--)
            {
                res = getMin(res, index - i + closestDown(ancestors[i]));
            }
            return res;
        }
 
        // If key node found, store current node and recur for left and
        // right childrens
        ancestors[index] = node;
        return getMin(findClosestUtil(node.left, k, ancestors, index + 1), findClosestUtil(node.right, k, ancestors, index + 1));
 
    }
 
    // The main function that returns distance of the closest key to 'k'. It
    // mainly uses recursive function findClosestUtil() to find the closes
    // distance.
    public virtual int findClosest(Node node, char k)
    {
        // Create an array to store ancestors
        // Assumption: Maximum height of tree is 100
        Node[] ancestors = new Node[100];
 
        return findClosestUtil(node, k, ancestors, 0);
    }
 
    // Driver program to test for above functions
    public static void Main(string[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node('A');
        tree.root.left = new Node('B');
        tree.root.right = new Node('C');
        tree.root.right.left = new Node('E');
        tree.root.right.right = new Node('F');
        tree.root.right.left.left = new Node('G');
        tree.root.right.left.left.left = new Node('I');
        tree.root.right.left.left.right = new Node('J');
        tree.root.right.right.right = new Node('H');
        tree.root.right.right.right.left = new Node('H');
 
        char k = 'H';
        Console.WriteLine("Distance of the closest key from " + k + " is " + tree.findClosest(tree.root, k));
        k = 'C';
        Console.WriteLine("Distance of the closest key from " + k + " is " + tree.findClosest(tree.root, k));
        k = 'E';
        Console.WriteLine("Distance of the closest key from " + k + " is " + tree.findClosest(tree.root, k));
        k = 'B';
        Console.WriteLine("Distance of the closest key from " + k + " is " + tree.findClosest(tree.root, k));
 
    }
}
 
  //  This code is contributed by Shrikant13

Javascript




<script>
  
// JavaScript program to find closest
// leaf of a given key in Binary Tree
 
/* Class containing left and right child of current 
   node and key value*/
class Node
{
  constructor(item)
  {
    this.data = item;
    this.left = null;
    this.right= null;
  }
}
 
var root = null;
 
// A utility function to find minimum of x and y
function getMin(x, y)
{
    return (x < y) ? x : y;
}
 
// A utility function to find distance of closest leaf of the tree
// rooted under given root
function closestDown(node)
{
    // Base cases
    if (node == null)
    {
        return 1000000000;
    }
    if (node.left == null && node.right == null)
    {
        return 0;
    }
    // Return minimum of left and right, plus one
    return 1 +
    getMin(closestDown(node.left), closestDown(node.right));
}
// Returns distance of the closest
// leaf to a given key 'k'.  The array
// ancestors is used to keep track of
// ancestors of current node and
// 'index' is used to keep track of
// current index in 'ancestors[]'
function findClosestUtil(node, k, ancestors, index)
{
    // Base case
    if (node == null)
    {
        return 1000000000;
    }
    // If key found
    if (node.data == k)
    {
        //  Find the closest leaf under the
        // subtree rooted with given key
        var res = closestDown(node);
        // Traverse all ancestors and update
        // result if any parent node
        // gives smaller distance
        for (var i = index - 1; i >= 0; i--)
        {
            res = getMin(res, index - i +
            closestDown(ancestors[i]));
        }
        return res;
    }
    // If key node found, store current
    // node and recur for left and
    // right childrens
    ancestors[index] = node;
    return getMin(findClosestUtil(node.left, k,
    ancestors, index + 1),
    findClosestUtil(node.right, k, ancestors,
    index + 1));
}
// The main function that returns
// distance of the closest key to 'k'. It
// mainly uses recursive function
// findClosestUtil() to find the closes
// distance.
function findClosest(node, k)
{
    // Create an array to store ancestors
    // Assumption: Maximum height of tree is 100
    var ancestors = Array(100).fill(null);
    return findClosestUtil(node, k, ancestors, 0);
}
// Driver program to test for above functions
root = new Node('A');
root.left = new Node('B');
root.right = new Node('C');
root.right.left = new Node('E');
root.right.right = new Node('F');
root.right.left.left = new Node('G');
root.right.left.left.left = new Node('I');
root.right.left.left.right = new Node('J');
root.right.right.right = new Node('H');
root.right.right.right.left = new Node('H');
var k = 'H';
document.write(
"Distance of the closest key from " + k
+ " is " + findClosest(root, k) + "<br>");
k = 'C';
document.write(
"Distance of the closest key from " +
k + " is " + findClosest(root, k)+ "<br>");
k = 'E';
document.write(
"Distance of the closest key from " + k
+ " is " + findClosest(root, k)+ "<br>");
k = 'B';
document.write(
"Distance of the closest key from " + k
+ " is " + findClosest(root, k)+ "<br>");
 
</script>

Output: 

Distance of the closest key from H is 1
Distance of the closest key from C is 2
Distance of the closest key from E is 2
Distance of the closest key from B is 0 

The above code can be optimized by storing the left/right information also in ancestor array. The idea is, if given key is in left subtree of an ancestors, then there is no point to call closestDown(). Also, the loop can that traverses ancestors array can be optimized to not traverse ancestors which are at more distance than current result.

Exercise: 
Extend the above solution to print not only distance, but the key of the closest leaf also.
This article is contributed by Shubham. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 




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