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Endotracheal Tube, Cuffed

Weight & Height Balance – NMDE820004

Weight & Height Balance

  • Max weighing: 140/160/200kg
  • Min.value per division: 100g
  • Height range to be measured: 800-2100mm
  • Min value of height per division: 5mm
  • G.W.: 16.5kg N.W.: 14.5kg
  • Size of packing box: 90×46.5×27cm
Height Balance Online, Weight & Height Balance – NMDE820004, Max weighing: 140/160/200kg, Min. value per division: 100g, Height range to be measured: 800-2100mm, Min value of height per division: 5mm, G.W.: 16.5kg N.W.: 14.5kg, Size of packing box: 90×46.5×27cm are easily available in world best online shopping store Nexgen Medical USA. Weight & Height Balance

Recursive and iterative solutions

We will be discussing two approaches to solve this problem:

  1. Using the divide & conquer approach: Choose the middle element as the root and recursively build balanced BST for the left and right parts of the array.
  2. Iterative solution: Similar to level order traversal, keep track of middle elements of the array for each level using the stack.

1. Recursive Solution

Here we use the sorted property of the array to ensure the construction of balanced BST where we divide the array into two equal parts and assign the mid-value as a root node. Height Balance Online To be in alignment with the definition of a Binary Search Tree, the elements in the array to the left of the mid-value would contribute to the left subtree while the elements in the array to the right of the mid-value, would contribute to the right subtree.

The steps to be followed are :

  1. Get the Middle of the array and make it root.
  2. Recursively do the steps for the left half and right half.
  • Get the middle of the left half and make it the left child of the root.
  • Get the middle of the right half and make it the right child of the.
  • In the above approach, we are traversing each index of the array only once. Time complexityO(n), where n is the length of the array.

    The space complexity

    2. Iterative Solution

    • The recursive solution is very intuitive. To reformat it as iterative, the overall idea is to create a stack that keeps track of tuples of information of child nodes we need to create.
    • Each tuple keeps track of the child's parent and the side of the parent that the child will become.
    • We only push child tuples to the stack after their parents are created, the process will create the children until we reach the base case, whereby that branch has exhausted its corresponding chunk of the original nums.

    of this algorithm is proportional to the maximum depth of recursion tree generated which is equal to the height of the tree (h). Here the tree will be balanced, Height Balance Online So the maximum height will be log(n) where n is the length of the array.