Tree
Depth-First Search
Breadth-First Search
Binary Search Tree
Binary Tree
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:
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 <= 105
Note: This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/
Solutions
Solution 1
Python3 Java C++ Go JavaScript
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20 # 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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37 /**
* 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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31 /**
* 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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31 /**
* 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
}
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15 var minDiffInBST = function ( root ) {
let ans = Number . MAX_SAFE_INTEGER ,
prev = Number . MAX_SAFE_INTEGER ;
const dfs = root => {
if ( ! root ) {
return ;
}
dfs ( root . left );
ans = Math . min ( ans , Math . abs ( root . val - prev ));
prev = root . val ;
dfs ( root . right );
};
dfs ( root );
return ans ;
};
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