LeetCode – Inorder Successor in BST II (Java)

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Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

The successor of a node p is the node with the smallest key greater than p.val. You will have direct access to the node but not to the root of the tree. Each node will have a reference to its parent node. A node is defined as the following:

// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node parent;
};
*/

Java Solution

public Node inorderSuccessor(Node x) {
  Node result = null;
 
  //case 1: right child is not null -> go down to get the next
  Node p = x.right;
  while(p!=null){ 
    result = p;
    p = p.left;
  }
 
  if(result != null){
    return result; 
  }
 
  //case 2: right child is null -> go up to the parent, 
  //until the node is a left child, return the parent
  p = x;
 
  while(p!=null){
    if(p.parent!=null && p.parent.left==p){
      return p.parent;
    } 
    p = p.parent;
  }
 
  return null;
}

If the tree is balanced, the time complexity is the height of the tree – O(log(n)). In the worst cast, the time is O(n). Space complexity is constant.

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