Given a binary tree and an integer **b** representing budget. The task is to find the count of maximum number of leaf nodes that can be visited under the given budget if **cost of visiting a leaf node** is equal to **level of that leaf node**.

**Note:** Root of the tree is at **level 1**.

**Examples:**

Input:b = 8 10 / \ 8 15 / / \ 3 11 18 \ 13Output:2 For the above binary tree, leaf nodes are 3, 13 and 18 at levels 3, 4 and 3 respectively. Cost for visiting leaf node 3 is 3 Cost for visiting leaf node 13 is 4 Cost for visiting leaf node 18 is 3 Thus with given budget = 8, we can at maximum visit two leaf nodes.Input:b = 1 8 / \ 7 10 / 3Output:0 For the above binary tree, leaf nodes are 3 and 10 at levels 3 and 2 respectively. Cost for visiting leaf node 3 is 3 Cost for visiting leaf node 10 is 2 In given budget = 1, we can't visit any leaf node.

**Approach:**

- Traverse the binary tree using level order traversal, and store the level of all the leaf nodes in a priority queue.
- Remove an element from the priority queue one by one, check if this value is within the budget.
- If
**Yes**then**subtract**this value from the budget and update**count = count + 1**. - Else, print the
**count of maximum leaf nodes**that can be visited within the given budget.

Below is the implementation of the above approach:

`// Java program to calculate the maximum number of leaf ` `// nodes that can be visited within the given budget ` `import` `java.io.*; ` `import` `java.util.*; ` `import` `java.lang.*; ` ` ` `// Class that represents a node of the tree ` `class` `Node { ` ` ` `int` `data; ` ` ` `Node left, right; ` ` ` ` ` `// Constructor to create a new tree node ` ` ` `Node(` `int` `key) ` ` ` `{ ` ` ` `data = key; ` ` ` `left = right = ` `null` `; ` ` ` `} ` `} ` ` ` `class` `GFG { ` ` ` ` ` `// Priority queue to store the levels ` ` ` `// of all the leaf nodes ` ` ` `static` `PriorityQueue<Integer> pq; ` ` ` ` ` `// Level order traversal of the binary tree ` ` ` `static` `void` `levelOrder(Node root) ` ` ` `{ ` ` ` `Queue<Node> q = ` `new` `LinkedList<>(); ` ` ` `int` `len, level = ` `0` `; ` ` ` `Node temp; ` ` ` ` ` `// If tree is empty ` ` ` `if` `(root == ` `null` `) ` ` ` `return` `; ` ` ` ` ` `q.add(root); ` ` ` ` ` `while` `(` `true` `) { ` ` ` ` ` `len = q.size(); ` ` ` `if` `(len == ` `0` `) ` ` ` `break` `; ` ` ` `level++; ` ` ` `while` `(len > ` `0` `) { ` ` ` ` ` `temp = q.remove(); ` ` ` ` ` `// If left child exists ` ` ` `if` `(temp.left != ` `null` `) ` ` ` `q.add(temp.left); ` ` ` ` ` `// If right child exists ` ` ` `if` `(temp.right != ` `null` `) ` ` ` `q.add(temp.right); ` ` ` ` ` `// If node is a leaf node ` ` ` `if` `(temp.left == ` `null` `&& temp.right == ` `null` `) ` ` ` `pq.add(level); ` ` ` `len--; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// Function to calculate the maximum number of leaf nodes ` ` ` `// that can be visited within the given budget ` ` ` `static` `int` `countLeafNodes(Node root, ` `int` `budget) ` ` ` `{ ` ` ` `pq = ` `new` `PriorityQueue<>(); ` ` ` `levelOrder(root); ` ` ` `int` `val; ` ` ` ` ` `// Variable to store the count of ` ` ` `// number of leaf nodes possible to visit ` ` ` `// within the given budget ` ` ` `int` `count = ` `0` `; ` ` ` ` ` `while` `(pq.size() != ` `0` `) { ` ` ` ` ` `// Removing element from ` ` ` `// min priority queue one by one ` ` ` `val = pq.poll(); ` ` ` ` ` `// If current val is under budget, the ` ` ` `// node can be visited ` ` ` `// Update the budget afterwards ` ` ` `if` `(val <= budget) { ` ` ` `count++; ` ` ` `budget -= val; ` ` ` `} ` ` ` `else` ` ` `break` `; ` ` ` `} ` ` ` `return` `count; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` ` ` `Node root = ` `new` `Node(` `10` `); ` ` ` `root.left = ` `new` `Node(` `8` `); ` ` ` `root.right = ` `new` `Node(` `15` `); ` ` ` `root.left.left = ` `new` `Node(` `3` `); ` ` ` `root.left.left.right = ` `new` `Node(` `13` `); ` ` ` `root.right.left = ` `new` `Node(` `11` `); ` ` ` `root.right.right = ` `new` `Node(` `18` `); ` ` ` ` ` `int` `budget = ` `8` `; ` ` ` ` ` `System.out.println(countLeafNodes(root, budget)); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

**Output:**

2

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