Description#
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:
![](https://spcdn.pages.dev/leetcode/problems/0783.Minimum%20Distance%20Between%20BST%20Nodes/images/bst1.jpg)
Input: root = [4,2,6,1,3]
Output: 1
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/0783.Minimum%20Distance%20Between%20BST%20Nodes/images/bst2.jpg)
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 <= 105
Note: This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/
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 minDiffInBST(self, root: Optional[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 minDiffInBST(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 minDiffInBST(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 minDiffInBST(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
}
|