C++ program to Replace a Node with Depth in a Binary Tree

Suppose we have a binary tree and we want to replace the depth of each node with its value. The depth of the node starts from 0 at the root node and increases by 1 for each level we go; for example, we have a binary tree like this;

Here we replace,

We do a simple preorder tree traversal and assign each node the depth.

Let’s look at some input scenarios −

Suppose we have values of nodes in binary tree as [3 2 7 5 4 6 9 7 2] and the output obtained from this input would be −

Input: [3 2 7 5 4 6 9 7 2]
Result:
Before: 6 5 9 2 4 3 7 7 2
After: 3 2 3 1 2 0 2 1 2

Suppose we have values of nodes in binary tree as [4 3 5 6 10 7 12 8 1] and the output obtained from this input would be −

Input: [4 3 5 6 10 7 12 8 1]
Result:
Before: 7 6 12 3 10 4 8 5 1
After: 3 2 3 1 2 0 2 1 2 

Example

Following is a C++ program to replace a node value in a binary tree with its corresponding depth within the tree −

#include <iostream>
using namespace std;
class Node {
   public:
   int value;
   Node *left, *right;
   Node(int value) {
      this->value = value;
      left = right = NULL;
   }
};
void solve(Node *node, int depth) {
   if (node == NULL) {
      return;
   }
   node->value = depth;
   solve(node->left, depth+1);
   solve(node->right, depth+1);
}
void printInorder(Node* root) {
   if (root == NULL) {
      return;
   }
   printInorder(root->left);
   cout << root->value <<" ";
   printInorder(root->right);
}
int main() {
   Node* root = new Node(1);
   root->left = new Node(2);
   root->right = new Node(3);
   root->left->left = new Node(4);
   root->left->right = new Node(5);
   root->left->left->left = new Node(8);
   root->left->left->right = new Node(9);
   root->right->left = new Node(6);
   root->right->right = new Node(7);
   cout << "Before: ";
   printInorder(root);
   solve(root, 0);
   cout << "\nAfter: ";
   printInorder(root);
   return 0;
}

Output

Before: 8 4 9 2 5 1 6 3 7
After: 3 2 3 1 2 0 2 1 2 

Conclusion

We conclude that using a simple preorder traversal of the tree and replacing the value with depth at each node is sufficient to solve the problem. There is an O(n) time complexity. Here is a article we hope will be helpful to you.

Prateek Jangid

Updated on: 10-Aug-2022

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