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# Implementing Backward Iterator in BST

• Difficulty Level : Easy
• Last Updated : 29 Jun, 2021

Given a Binary search tree, the task is to implement backward iterator on it with the following functions.

1. curr(): returns the pointer to current element.
2. prev(): iterates to the previous largest element in the Binary Search Tree.
3. isEnd(): returns true if there’s no node left to traverse else false.

Iterator traverses the BST in decreasing order. We will implement the iterator using a stack data structure.

Initialisation:

• We will create a stack named “q” to store the nodes of BST.
• Create a variable “curr” and initialise it with pointer to root.
• While “curr” is not NULL
• Push “curr” in the stack ‘q’.
• Set curr = curr -> right

curr()

Returns the value at the top of the stack ‘q’.
Note: It might throw segmentation fault if the stack is empty.

Time Complexity: O(1)

prev()

• Declare pointer variable “curr” which points to node.
• Set curr = q.top()->left.
• Pop top most element of stack.
• While “curr” is not NULL
• Push “curr” in the stack ‘q’.
• Set curr = curr -> right.

Time Complexity: O(1) on average of all calls. Can be O(h) for a single call in the worst case.

isEnd()

Returns true if stack “q” is empty else returns false.

Time Complexity: O(1)
Worst Case space complexity for this implementation of iterators is O(h). It should be noticed that
iterator points to the top-most element of the stack.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Node of the binary tree``struct` `node {``    ``int` `data;``    ``node* left;``    ``node* right;``    ``node(``int` `data)``    ``{``        ``this``->data = data;``        ``left = NULL;``        ``right = NULL;``    ``}``};` `// Iterator for BST``class` `bstit {``private``:``    ``// Stack to store the nodes``    ``// of BST``    ``stack q;` `public``:``    ``// Constructor for the class``    ``bstit(node* root)``    ``{``        ``// Initializing stack``        ``node* curr = root;``        ``while` `(curr != NULL)``            ``q.push(curr), curr = curr->right;``    ``}` `    ``// Function to return``    ``// current element iterator``    ``// is pointing to``    ``node* curr()``    ``{``        ``return` `q.top();``    ``}` `    ``// Function to iterate to previous``    ``// element of BST``    ``void` `prev()``    ``{``        ``node* curr = q.top()->left;``        ``q.pop();``        ``while` `(curr != NULL)``            ``q.push(curr), curr = curr->right;``    ``}` `    ``// Function to check if``    ``// stack is empty``    ``bool` `isEnd()``    ``{``        ``return` `!(q.size());``    ``}``};` `// Function to test the iterator``void` `iterate(bstit it)``{``    ``while` `(!it.isEnd())``        ``cout << it.curr()->data << ``" "``, it.prev();``}` `// Driver code``int` `main()``{``    ``node* root = ``new` `node(5);``    ``root->left = ``new` `node(3);``    ``root->right = ``new` `node(7);``    ``root->left->left = ``new` `node(2);``    ``root->left->right = ``new` `node(4);``    ``root->right->left = ``new` `node(6);``    ``root->right->right = ``new` `node(8);` `    ``// Iterator to BST``    ``bstit it(root);` `    ``// Function call to test the iterator``    ``iterate(it);` `    ``return` `0;``}`

## Javascript

 ``
Output:
`8 7 6 5 4 3 2`

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