Inorder Successor in BST II - Problem
You're given a node in a Binary Search Tree (BST) and need to find its inorder successor - the next node that would appear in an inorder traversal of the tree.
The inorder successor of a node is the node with the smallest value greater than the current node's value. If no such node exists, return null.
The challenge: You only have access to the given node, not the root of the tree! However, each node has a reference to its parent, which makes this problem solvable through clever tree navigation.
Node Structure:
class Node {
public int val;
public Node left;
public Node right;
public Node parent;
}Goal: Navigate the tree structure using parent pointers to find the successor without needing the root reference.
Input & Output
basic_successor.py — Python
$
Input:
node = Node(4) with tree structure [2,1,4,null,null,3,6]
›
Output:
Node(6)
💡 Note:
Node 4 has a right subtree starting with 6. Since 6 has no left child, 6 is the leftmost node in the right subtree and thus the successor.
no_right_subtree.py — Python
$
Input:
node = Node(6) with tree structure [2,1,4,null,null,3,6]
›
Output:
null
💡 Note:
Node 6 has no right subtree. Going up the parent chain, 6 is always the right child of its ancestors, so there's no successor.
left_child_case.py — Python
$
Input:
node = Node(3) with tree structure [5,3,8,2,4,6,9]
›
Output:
Node(4)
💡 Note:
Node 3 has a right subtree. The leftmost node in the right subtree starting from 4 is 4 itself, making it the successor.
Visualization
Tap to expand
Understanding the Visualization
1
Check Right Path
If there's a right subtree, the successor is the leftmost node there
2
Navigate Up
Otherwise, go up until you find an ancestor where you came from the left
3
Return Result
That ancestor is your successor, or null if none exists
Key Takeaway
🎯 Key Insight: BST structure + parent pointers enable O(h) successor finding through smart navigation rather than full traversal
Time & Space Complexity
Time Complexity
O(n)
Need to visit all n nodes during inorder traversal
✓ Linear Growth
Space Complexity
O(n)
Store all nodes in array plus recursion stack depth
⚡ Linearithmic Space
Constraints
- The number of nodes in the tree is in the range [1, 104]
- -105 ≤ Node.val ≤ 105
- All Node.val are unique
- node is guaranteed to be in the tree
- Each node has a reference to its parent node
💡
Explanation
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