Description#
Given the root
of a Binary Search Tree (BST), return the minimum absolute 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, 104]
. 0 <= Node.val <= 105
Note: This question is the same as 783: https://leetcode.com/problems/minimum-distance-between-bst-nodes/
Solutions#
Solution 1#
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| # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def getMinimumDifference(self, root: TreeNode) -> int:
def dfs(root):
if root is None:
return
dfs(root.left)
nonlocal ans, prev
ans = min(ans, abs(prev - root.val))
prev = root.val
dfs(root.right)
ans = prev = inf
dfs(root)
return ans
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| /**
* 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 {
private int ans;
private int prev;
private int inf = Integer.MAX_VALUE;
public int getMinimumDifference(TreeNode root) {
ans = inf;
prev = inf;
dfs(root);
return ans;
}
private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
ans = Math.min(ans, Math.abs(root.val - prev));
prev = root.val;
dfs(root.right);
}
}
|
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| /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
const int inf = INT_MAX;
int ans;
int prev;
int getMinimumDifference(TreeNode* root) {
ans = inf, prev = inf;
dfs(root);
return ans;
}
void dfs(TreeNode* root) {
if (!root) return;
dfs(root->left);
ans = min(ans, abs(prev - root->val));
prev = root->val;
dfs(root->right);
}
};
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| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func getMinimumDifference(root *TreeNode) int {
inf := 0x3f3f3f3f
ans, prev := inf, inf
var dfs func(*TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
dfs(root.Left)
ans = min(ans, abs(prev-root.Val))
prev = root.Val
dfs(root.Right)
}
dfs(root)
return ans
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
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| // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
#[allow(dead_code)]
pub fn get_minimum_difference(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut ret = i32::MAX;
let mut prev = i32::MAX;
Self::traverse(root, &mut prev, &mut ret);
ret
}
#[allow(dead_code)]
fn traverse(root: Option<Rc<RefCell<TreeNode>>>, prev: &mut i32, ans: &mut i32) {
let left = root.as_ref().unwrap().borrow().left.clone();
let right = root.as_ref().unwrap().borrow().right.clone();
let val = root.as_ref().unwrap().borrow().val;
if !left.is_none() {
Self::traverse(left.clone(), prev, ans);
}
*ans = std::cmp::min(*ans, (*prev - val).abs());
*prev = val;
if !right.is_none() {
Self::traverse(right.clone(), prev, ans);
}
}
}
|