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DSA Javaprogramming~5 mins

BST Inorder Successor in DSA Java - Cheat Sheet & Quick Revision

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Recall & Review
beginner
What is the inorder successor of a node in a Binary Search Tree (BST)?
The inorder successor of a node in a BST is the node with the smallest key greater than the given node's key. It is the next node visited in an inorder traversal.
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beginner
How do you find the inorder successor if the node has a right child?
If the node has a right child, the inorder successor is the leftmost node in the node's right subtree.
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intermediate
How do you find the inorder successor if the node has no right child?
If the node has no right child, the inorder successor is one of its ancestors. Specifically, it is the lowest ancestor of the node whose left child is also an ancestor of the node.
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intermediate
What is the time complexity of finding the inorder successor in a BST?
The time complexity is O(h), where h is the height of the BST, because you may need to traverse up or down the tree along a path.
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beginner
In an inorder traversal, what is the order of nodes visited?
Inorder traversal visits nodes in ascending order: left subtree, current node, then right subtree.
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What is the inorder successor of a node with a right subtree?
AThe right child itself
BThe leftmost node in the right subtree
CThe parent node
DThe rightmost node in the left subtree
If a node has no right child, how do you find its inorder successor?
AThere is no inorder successor
BGo to the right child
CGo to the left child
DGo up to the lowest ancestor whose left child is also an ancestor of the node
What is the inorder successor of the maximum node in a BST?
ANo successor (null)
BThe root node
CThe minimum node
DThe parent node
What traversal order does inorder successor relate to?
AInorder
BPreorder
CPostorder
DLevel order
What is the worst-case time complexity to find the inorder successor in a BST?
AO(n)
BO(log n)
CO(h)
DO(1)
Explain how to find the inorder successor of a node in a BST when the node has a right child and when it does not.
Think about the next bigger value in the sorted order of BST nodes.
You got /2 concepts.
    Describe the relationship between inorder traversal and inorder successor in a BST.
    Consider how nodes are ordered in inorder traversal.
    You got /2 concepts.

      Practice

      (1/5)
      1. What is the inorder successor of a node in a Binary Search Tree (BST)?
      easy
      A. The node with the smallest key greater than the given node's key
      B. The node with the largest key smaller than the given node's key
      C. The root node of the BST
      D. The left child of the given node

      Solution

      1. Step 1: Understand inorder traversal order

        Inorder traversal visits nodes in ascending order of their keys in a BST.
      2. Step 2: Define inorder successor

        The inorder successor of a node is the next node visited after it in inorder traversal, which is the smallest node greater than it.
      3. Final Answer:

        The node with the smallest key greater than the given node's key -> Option A
      4. Quick Check:

        Inorder successor = smallest greater key [OK]
      Hint: Inorder successor is next bigger node in sorted order [OK]
      Common Mistakes:
      • Confusing successor with predecessor
      • Thinking successor is always right child
      • Assuming successor is parent node
      2. Which of the following Java method signatures correctly represents a function to find the inorder successor in a BST node class?
      easy
      A. public void inorderSuccessor(Node p)
      B. public Node inorderSuccessor(Node root, Node p)
      C. public int inorderSuccessor(Node root)
      D. public Node inorderSuccessor(int key)

      Solution

      1. Step 1: Identify required inputs and outputs

        To find inorder successor, we need the BST root and the target node; output is the successor node.
      2. Step 2: Match method signature

        public Node inorderSuccessor(Node root, Node p) takes root and node, returns Node, matching requirements.
      3. Final Answer:

        public Node inorderSuccessor(Node root, Node p) -> Option B
      4. Quick Check:

        Input root and node, output node [OK]
      Hint: Method needs root and target node, returns successor node [OK]
      Common Mistakes:
      • Missing root parameter
      • Returning int instead of Node
      • Using void return type
      3. Given the BST below and the node with value 15, what is the inorder successor?
             20
            /  \
          10    30
            \
            15
      
      medium
      A. 20
      B. 15
      C. 30
      D. 10

      Solution

      1. Step 1: Locate node 15 and check right subtree

        Node 15 has no right child, so successor is ancestor with greater value.
      2. Step 2: Find lowest ancestor greater than 15

        Ancestor 20 is the next bigger node after 15.
      3. Final Answer:

        20 -> Option A
      4. Quick Check:

        Next bigger after 15 is 20 [OK]
      Hint: If no right child, successor is nearest bigger ancestor [OK]
      Common Mistakes:
      • Choosing 30 which is too big
      • Choosing 10 which is smaller
      • Choosing the node itself
      4. Consider this buggy Java code snippet to find the inorder successor of a node in BST:
      public Node inorderSuccessor(Node root, Node p) {
          if (p.right != null) {
              Node curr = p.right;
              while (curr.left != null) {
                  curr = curr.left;
              }
              return curr;
          }
          Node succ = null;
          while (root != null) {
              if (p.val < root.val) {
                  succ = root;
                  root = root.left;
              } else {
                  root = root.right;
              }
          }
          return succ;
      }
      What is the main bug in this code?
      medium
      A. It uses wrong comparison operator in while loop
      B. It incorrectly updates successor when p.val >= root.val
      C. It does not handle the case when p has no right child
      D. It may return null even if successor exists

      Solution

      1. Step 1: Analyze right subtree handling

        If p has right child, code correctly finds leftmost node in right subtree.
      2. Step 2: Analyze ancestor search loop

        Loop updates successor when p.val < root.val, else moves right. This is correct logic.
      3. Step 3: Identify potential null return

        If p is the largest node, no successor exists, so null return is valid. Otherwise, code works.
      4. Final Answer:

        It may return null even if successor exists -> Option D
      5. Quick Check:

        Null return possible if logic misses successor [OK]
      Hint: Check if successor can be missed when traversing ancestors [OK]
      Common Mistakes:
      • Assuming null means no successor always
      • Ignoring right subtree case
      • Misreading comparison operators
      5. You have a BST and want to find the inorder successor of node with value 25. The BST is:
              40
             /  \
           20    60
          /  \  /  \
        10  30 50  70
             \
             25
      
      Which node is the inorder successor of 25 and why?
      hard
      A. 30, because it is the smallest node greater than 25 in the right subtree
      B. 50, because it is the leftmost node in the right subtree of 25
      C. 40, because 25 has no right child and 40 is the nearest bigger ancestor
      D. 20, because it is the parent of 25

      Solution

      1. Step 1: Check if node 25 has right child

        Node 25 has no right child, so successor is ancestor or next bigger node.
      2. Step 2: Find nearest ancestor greater than 25

        Node 40 is the nearest ancestor greater than 25, so it is the inorder successor.
      3. Final Answer:

        40, because 25 has no right child and 40 is the nearest bigger ancestor -> Option C
      4. Quick Check:

        Next bigger node after 25 is 40 [OK]
      Hint: Successor is leftmost node in right subtree or nearest bigger ancestor [OK]
      Common Mistakes:
      • Choosing 30 which is smaller than 25
      • Choosing 50 which is too far
      • Choosing parent 20 which is smaller
      Recommended

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