Inorder Predecessor of Node in BST
https://workat.tech/problem-solving/practice/inorder-predecessor-bst
The inorder predecessor of a node p is the node q that comes just before p in the binary tree's inorder traversal.
Given the root node of a binary search tree and the node p, find the inorder predecessor of node p. If it does not exist, return null.
Testing
Input Format
The first line contains an integer T denoting the number of test cases.
For each test case, the input has 2 lines:
- The first line contains an integer n denoting the number of nodes in the tree (including the NULL nodes).
- The second line contains n space-separated integers that will form the binary tree. The integers follow level order traversal of the tree where -1 indicates a NULL node.
- The third line contains an integer denoting the 0-based index of p in the above list.
Output Format
For each test case, the output contains an integer with the value of the inorder predecessor. In case the predecessor doesn't exist the value is -1.
Sample Input
4
9
2 1 3 -1 -1 -1 5 4 7
2
7
6 3 21 -1 -1 -1 89
1
12
8 3 9 -1 4 -1 10 -1 -1 -1 12 11
11
4
28 14 -1 11
1Expected Output
2
-1
10
111 <= T <= 101 <= n <= 1051 <= node value <= 109
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| class Solution { | |
| /* This is the Node class definition | |
| class Node { | |
| public Node left; | |
| public Node right; | |
| public int data; | |
| public Node(int data) { | |
| this.data = data; | |
| } | |
| } | |
| */ | |
| Node findPredecessor(Node root, Node p) { | |
| Node ans = null; | |
| Node node = root; | |
| while (node != null) { | |
| if (node.data < p.data) { | |
| ans = node; | |
| node = node.right; | |
| } else { | |
| node = node.left; | |
| } | |
| } | |
| return ans; | |
| } | |
| } |
