Level order sorted binary tree from a binary tree

algorithmsbinary treesorting

Lets suppose we have a binary tree . Tree node structure is as 

struct node 
{
   int val ;
   struct node *left , *right ;
}

Now we have to sort the tree in level order manner . For example lets suppose we have original tree like this 

root = 6
root->left = 1 
root->right = 9
root->left->left = 8
root->right->right = 2

Then after applying level order sorting , our tree will be 

root = 1
root->left = 2
root->right = 6
root->left->left = 8
root->right->right = 9

One way of doing this problem is just take all the values in a array and apply any of the standard sorting algorithm and then put back the values in tree in level order traversal manner. This solution is not efficient in term of space and time both.

What algorithm can I use to decrease operations and memory usage from my proposed plan? Is there a more efficient algorithm to do this without the use of a sorted array?

Level order sorting :
Level order sorting means if you do level order traversal of the tree then you should get a sorted non-decreasing sequence .

Constraint :
Shape of the tree should not alter that is if in original tree left child of the root does not have right child (as in my example) then after sorting too, it should not has right child .

Best Answer

Sounds like you need a type of heap with some modified properties. First, use a min-heap instead of a max-heap: every node is less than the values of its children, and add a constraint that a node is greater than its siblings to the left.

One of the neat features of a heap is that it is a balanced binary tree: the depth of all leaf nodes vary by at most 1. Furthermore, leaf nodes are populated left to right. While this is not quite the same representation as you have, it makes the whole thing a ton easier.

If you take a tree with these properties and perform a level-order traversal, you can easily convert it to and from an array:

root = 6
root->left = 1 
root->right = 9
root->left->left = 8
root->left->right = 2

Level-order traversal gives the following: 6, 1, 9, 8, 2.

Note how each level takes 2^n elements in that array, where n is the depth of the nodes at that level. The root node has depth 0, so it takes 2^0 = 1 element. Next level has depth 1, so it takes 2^1 = 2 elements. This property means there is a direct correlation between a specific element in the tree and its array position.

Now perform an efficient sort on that array, and you get: 1, 2, 6, 8, 9.

Perform the reverse of the first transformation. Element 0 is the root, Elements 1 and 2 are the next level, elements 3, 4, (and 5, 6 if present) are the next level.

root = 1
root->left = 2
root->right = 6
root->left->left = 8
root->left->right = 9

If you need to maintain the fact that leaves are not populated left to right, or the tree is not balanced, you can add gaps by saying certain nodes are null and modify a sort algorithm to leave null values in-place. Not ideal, but it can allow you to leverage the properties of a heap to do this. Level-order sorting pretty much has to be done using a heap as far as I am aware (made it through a B.S. and M.S. in computer science without ever hearing or seeing another way to do this, and you know the real world doesn't care).

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