Your solution calls both isBalanced and height, always. For every single node in the tree. The faster solution calls treeDepth for every single node, but bails out early and doesn't call isBalanced if it knows the tree is not balanced. It's an optimization to not call unnecessary (recursive/expensive) functions....

c,algorithm,binary-tree,tree-balancing

The basic idea is as follows. For insertions, you first insert your new node at a leaf exactly as you would for a non-balanced tree. Then you work your way up the tree towards the root, making sure that, for each node, the difference in height between the left and...

As you can read on wikipedia In an AVL tree, the heights of the two child subtrees of any node differ by at most one That doesn't imply that you have a complete tree ! See examples of balances tree in the wikipedia page linked. So for any of the...

algorithm,data-structures,b-tree,avl-tree,tree-balancing

The answer is... yes and no. B-trees don't need to perform rotations because they have some slack with how many different keys they can pack into a node. As you add more and more keys into a B-tree, you can avoid the tree becoming lopsided by absorbing those keys into...

c++,data-structures,set,binary-search-tree,tree-balancing

The standard doesn't say how the containers are to be implemented, so you can't count on an RB or a AVL tree. In practice... the complexity constraints are such that I don't know of any other implementations which would fit the bill. But it's in the complexity constraints that you...

algorithm,data-structures,binary-search-tree,red-black-tree,tree-balancing

Re-balancing may make a sibling of a node its new parent, but it can not change the relative order. Keep in mind that the red-black tree is a binary search tree and thus it should keep elements smaller than a given element in its left subtree and elements greater than...

java,recursion,tree,tree-balancing

This is easier solved with a dedicated class that represents an expression, and not with an array. Then you can simply enumerate all possible trees. A mix of another answer that I wrote for a similar task, and an answer that shows how to generate all binary trees gave this:...