Problem
Given the root of a Binary Search Tree (BST), return the minimum difference between the values of any two different nodes in the tree.
Example 1:
Input: root = [4,2,6,1,3]
Output: 1
Example 2:
Input: root = [1,0,48,null,null,12,49]
Output: 1
Constraints:
The number of nodes in the tree is in the range
[2, 100].0 <= Node.val <= 10^5
Note: This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/
Solution (Java)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
ArrayList<Integer> l = new ArrayList<>();
public void inorder(TreeNode root){
if(root == null){
return;
}
inorder(root.left);
l.add(root.val);
inorder(root.right);
}
public int minDiffInBST(TreeNode root) {
inorder(root);
int min = 10000;
for(int i=1;i<l.size();i++){
min = Math.min(min,l.get(i)-l.get(i-1));
}
return min;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).