MUST READ ** inorder BST gives sorted list *** preorder(I think any traversal should work) and Treeset ** when it's not a BST Recursive .. must read!! Time Complexity - O(n) Space Complexity - O(1)
From https://leetcode.com/problems/minimum-absolute-difference-in-bst/#/description
// The idea is to put values in a TreeSet and then every time we can use O(lgN) time to lookup for the nearest values.
// Solution 2 - Pre-Order traverse, time complexity O(NlgN), space complexity O(N).
public class Solution {
TreeSet<Integer> set = new TreeSet<>();
int min = Integer.MAX_VALUE;
public int getMinimumDifference(TreeNode root) {
if (root == null) return min;
if (!set.isEmpty()) {
if (set.floor(root.val) != null) {
min = Math.min(min, root.val - set.floor(root.val));
}
if (set.ceiling(root.val) != null) {
min = Math.min(min, set.ceiling(root.val) - root.val);
}
}
set.add(root.val);
getMinimumDifference(root.left);
getMinimumDifference(root.right);
return min;
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int min = Integer.MAX_VALUE;
Integer prev = null;
public int getMinimumDifference(TreeNode root) {
if (root == null){
return 0; // any int value is fine
}
getMinimumDifference(root.left);
if(prev != null){
min = Math.min(min, root.val - prev);
}
prev = root.val;
getMinimumDifference(root.right);
return min;
}
}