Given a Binary Tree consisting of **N** nodes, the task is to print the Level Order Traversal after replacing the value of each node with its nearest power of the minimum value of the previous level in the original tree.**Note:** For any case of two nearest powers, select the maximum among them.

**Examples:**

Input:7

/ \

4 11

/

23Output:7 4 11 23 NExplanation:

- Node value at level 0 remains unchanged, i.e. 7.
- Power of 7 nearest to 4 is 7
^{1}= 7.

Power of 7 nearest to 11 is 7^{1}= 7.

Therefore, nodes at level 1 becomes {7, 7}.- Minimum node value at level 1 is 4.

Power of 4 nearest to 23 is 4^{4}= 16.

Therefore, node at level 2 becomes {16}.The resultant tree after completing the above operations is as follows:

7

/ \

4 11

/

23

Input:3

/ \

2 6

/ \ \

45 71 25Output:3 3 9 32 64 N 32

**Approach: **The idea is to perform the Level Order Traversal using a Queue to solve the problem.

Follow the steps below to solve the problem:

- Define a function, say
**nearestPow(X, Y),**to find the nearest power of an integer**Y**:- Find
**log(X) base Y**and store it in a variable, say**K**. - Return
**Y**if^{K}**abs(X – Y**is less than^{K})**abs(Y**. Otherwise, return^{(K + 1)}– X)**Y**.^{(K + 1)}

- Find
- Initialize two variables, say
**minCurr**and**minPrev,**to store the minimum value of the current level and the minimum value of the previous level respectively. - Initially assign
**minPrev = root.val**and initialize a queue, say**Q**to store the nodes for level order traversal. - Iterate while
**Q**is not empty():- Store the first node of the queue in a variable, say
**temp**, and delete the first node from queue**Q**. - Assign the value of
**minCurr**to**minPrev**and update**minCurr = 10**.^{18} - Iterate over the range
**[0, length(Q) – 1]**and update the**minCurr**as**minCurr = min(minCurr, temp.val)**and assign the nearest power of the integer**minPrev**to**temp.val**. - In each iteration of the above step push the
**temp.left**and**temp.right**if the respective nodes are not**NULL**.

- Store the first node of the queue in a variable, say
- After completing the above steps, print the level order traversal of the updated Tree.

Below is the implementation of the above approach:

## Python3

`# Python program for the above approach` `import` `math` ` ` `# Structure of a Node of a Tree` `class` `TreeNode:` ` ` `def` `__init__(` `self` `, val` `=` `0` `, left` `=` `None` `, right` `=` `None` `):` ` ` `self` `.val ` `=` `val` ` ` `self` `.left ` `=` `left` ` ` `self` `.right ` `=` `right` ` ` ` ` `# Function to calculate the` `# nearest power of an integer` `def` `nearestPow(x, base):` ` ` `k ` `=` `int` `(math.log(x, base))` ` ` `if` `abs` `(base` `*` `*` `k ` `-` `x) < ` `abs` `(base` `*` `*` `(k` `+` `1` `) ` `-` `x):` ` ` `return` `base` `*` `*` `k` ` ` `else` `:` ` ` `return` `base` `*` `*` `(k` `+` `1` `)` ` ` `# Iterative method to perform` `# Level Order Traversal` `def` `printLevelOrder(root):` ` ` ` ` `# Base Case` ` ` `if` `root ` `is` `None` `:` ` ` `return` ` ` ` ` `# Queue for Level` ` ` `# Order Traversal` ` ` `q ` `=` `[]` ` ` ` ` `# Enqueue root` ` ` `q.append(root)` ` ` ` ` `while` `q:` ` ` ` ` `# Stores number of` ` ` `# nodes at current level` ` ` `count ` `=` `len` `(q)` ` ` ` ` `# Dequeue all nodes of the current` ` ` `# level and Enqueue all nodes of` ` ` `# the next level` ` ` `while` `count > ` `0` `:` ` ` `temp ` `=` `q.pop(` `0` `)` ` ` `print` `(temp.val, end` `=` `' '` `)` ` ` ` ` `# Push the left subtree` ` ` `# if not empty` ` ` `if` `temp.left:` ` ` `q.append(temp.left)` ` ` ` ` `# Push the right subtree` ` ` `# if not empty` ` ` `if` `temp.right:` ` ` `q.append(temp.right)` ` ` ` ` `# Decrement count by 1` ` ` `count ` `-` `=` `1` ` ` ` ` `# Function to replace each node` `# with nearest power of minimum` `# value of previous level` `def` `replaceNodes(root):` ` ` ` ` `# Stores the nodes of tree to` ` ` `# traverse in level order` ` ` `que ` `=` `[root]` ` ` ` ` `# Stores current level` ` ` `lvl ` `=` `1` ` ` ` ` `# Stores the minimum` ` ` `# value of previous level` ` ` `minPrev ` `=` `root.val` ` ` ` ` `# Stores the minimum` ` ` `# value of current level` ` ` `minCurr ` `=` `root.val` ` ` ` ` `# Iterate while True` ` ` `while` `True` `:` ` ` ` ` `# Stores length of queue` ` ` `length ` `=` `len` `(que)` ` ` ` ` `# If length is zero` ` ` `if` `not` `length:` ` ` `break` ` ` ` ` `# Assign minPrev = minCurr` ` ` `minPrev ` `=` `minCurr` ` ` `minCurr ` `=` `1000000000000000000` ` ` ` ` `# Iterate over range [0, length - 1]` ` ` `while` `length:` ` ` ` ` `# Stores current node of tree` ` ` `temp ` `=` `que.pop(` `0` `)` ` ` ` ` `# Update minCurr` ` ` `minCurr ` `=` `min` `(temp.val, minCurr)` ` ` ` ` `# Replace current node with` ` ` `# nearest power of minPrev` ` ` `temp.val ` `=` `nearestPow(temp.val, minPrev)` ` ` ` ` `# Left child is not Null` ` ` `if` `temp.left:` ` ` ` ` `# Append temp.left node` ` ` `# in the queue` ` ` `que.append(temp.left)` ` ` ` ` `# If right child is not Null` ` ` `if` `temp.right:` ` ` ` ` `# Append temp.right node` ` ` `# in the queue` ` ` `que.append(temp.right)` ` ` ` ` `# Decrement length by one` ` ` `length ` `-` `=` `1` ` ` ` ` `# Increment level by one` ` ` `lvl ` `+` `=` `1` ` ` ` ` `# Function Call to perform the` ` ` `# Level Order Traversal` ` ` `printLevelOrder(root)` ` ` ` ` `# Driver Code` ` ` `# Given Tree` `root ` `=` `TreeNode(` `7` `)` `root.left ` `=` `TreeNode(` `4` `)` `root.right ` `=` `TreeNode(` `11` `)` `root.left.right ` `=` `TreeNode(` `23` `)` ` ` `replaceNodes(root)` |

**Output:**

7 7 7 16

**Time Complexity: **O(N)**Auxiliary Space: **O(N)

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the **Essential Maths for CP Course** at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**