In Order Successor in Binary Search Tree

Question

Given a node in a BST, how does one find the next higher key?

Answer 1

We can find the successor in O(log n) without using parent pointers (for a balanced tree).

The idea is very similar to when you have parent pointers.

We can define a recursive function that achieves this as follows:

• If the current node is the target, return the left-most / smallest node of its right subtree, if it exists.
• Recurse left if the target is smaller than the current node, and right if it's greater.
• If the target is to the left and we haven't found a successor yet, return the current node.

Pseudo-code:

Key successor(Node current, Key target):
   if current == null
      return null
   if target == current.key
      if current.right != null
         return leftMost(current.right).key
      else
         return specialKey
   else
      if target < current.key
         s = successor(current.left, target)
         if s == specialKey
            return current.key
         else
            return s
      else
         return successor(current.right, target)

Node leftMost(Node current):
    while current.left != null
       current = current.left
    return current

Live Java demo.