Given root of a binary tree and a key x in it, find distance of the given key from root. Distance means number of edges between two nodes.
Examples:
Input : x = 45,
Root of below tree
5
/ \
10 15
/ \ / \
20 25 30 35
\
45
Output : Distance = 3
There are three edges on path
from root to 45.
For more understanding of question,
in above tree distance of 35 is two
and distance of 10 is 1.
The idea is to traverse the tree from root. Check if x is present at root or in left subtree or in right subtree. We initialize distance as -1 and add 1 to distance for all three cases.
C++
// C++ program to find distance of a given // node from root. #include <bits/stdc++.h> using namespace std; // A Binary Tree Node struct Node { int data; Node *left, *right; }; // A utility function to create a new Binary // Tree Node Node *newNode(int item) { Node *temp = new Node; temp->data = item; temp->left = temp->right = NULL; return temp; } // Returns -1 if x doesn't exist in tree. Else // returns distance of x from root int findDistance(Node *root, int x) { // Base case if (root == NULL) return -1; // Initialize distance int dist = -1; // Check if x is present at root or in left // subtree or right subtree. if ((root->data == x) || (dist = findDistance(root->left, x)) >= 0 || (dist = findDistance(root->right, x)) >= 0) return dist + 1; return dist; } // Driver Program to test above functions int main() { Node *root = newNode(5); root->left = newNode(10); root->right = newNode(15); root->left->left = newNode(20); root->left->right = newNode(25); root->left->right->right = newNode(45); root->right->left = newNode(30); root->right->right = newNode(35); cout << findDistance(root, 45); return 0; } |
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Java
// Java program to find distance of a given // node from root. import java.util.*; class GfG { // A Binary Tree Node static class Node { int data; Node left, right; } // A utility function to create a new Binary // Tree Node static Node newNode(int item) { Node temp = new Node(); temp.data = item; temp.left = null; temp.right = null; return temp; } // Returns -1 if x doesn't exist in tree. Else // returns distance of x from root static int findDistance(Node root, int x) { // Base case if (root == null) return -1; // Initialize distance int dist = -1; // Check if x is present at root or in left // subtree or right subtree. if ((root.data == x) || (dist = findDistance(root.left, x)) >= 0 || (dist = findDistance(root.right, x)) >= 0) return dist + 1; return dist; } // Driver Program to test above functions public static void main(String[] args) { Node root = newNode(5); root.left = newNode(10); root.right = newNode(15); root.left.left = newNode(20); root.left.right = newNode(25); root.left.right.right = newNode(45); root.right.left = newNode(30); root.right.right = newNode(35); System.out.println(findDistance(root, 45)); } } |
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Python3
# Python3 program to find distance of # a given node from root. # A class to create a new Binary # Tree Node class newNode: def __init__(self, item): self.data = item self.left = self.right = None # Returns -1 if x doesn't exist in tree. # Else returns distance of x from root def findDistance(root, x): # Base case if (root == None): return -1 # Initialize distance dist = -1 # Check if x is present at root or # in left subtree or right subtree. if (root.data == x): return dist + 1 else: dist = findDistance(root.left, x) if dist >= 0: return dist + 1 else: dist = findDistance(root.right, x) if dist >= 0: return dist + 1 return dist # Driver Code if __name__ == '__main__': root = newNode(5) root.left = newNode(10) root.right = newNode(15) root.left.left = newNode(20) root.left.right = newNode(25) root.left.right.right = newNode(45) root.right.left = newNode(30) root.right.right = newNode(35) print(findDistance(root, 45)) # This code is contributed by PranchalK |
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C#
// C# program to find distance of a given // node from root. using System; class GfG { // A Binary Tree Node class Node { public int data; public Node left, right; } // A utility function to create // a new Binary Tree Node static Node newNode(int item) { Node temp = new Node(); temp.data = item; temp.left = null; temp.right = null; return temp; } // Returns -1 if x doesn't exist in tree. Else // returns distance of x from root static int findDistance(Node root, int x) { // Base case if (root == null) return -1; // Initialize distance int dist = -1; // Check if x is present at root or in left // subtree or right subtree. if ((root.data == x) || (dist = findDistance(root.left, x)) >= 0 || (dist = findDistance(root.right, x)) >= 0) return dist + 1; return dist; } // Driver code public static void Main(String[] args) { Node root = newNode(5); root.left = newNode(10); root.right = newNode(15); root.left.left = newNode(20); root.left.right = newNode(25); root.left.right.right = newNode(45); root.right.left = newNode(30); root.right.right = newNode(35); Console.WriteLine(findDistance(root, 45)); } } // This code is contributed by 29AjayKumar |
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Output:
3
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