Usual approach using linear space (stack) We can easily do the traversal by using a stack where we keep pushing left. If no left is there then we pop an element to print it and then push its right. This is O(n) time and O(n) space algorithm. public static void InorderTraversal(BTNode root){ Stack<BTNode> stack […]
Tag Archives: inorder
From wiki, A binary search tree is a rooted binary tree, whose internal nodes each store a key (and optionally, an associated value) and each have two distinguished sub-trees, commonly denoted left and right. The tree additionally satisfies the binary search tree property, which states that the key in each node must be greater than all […]
