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: Inorder Traversal
The problem requires us to find the minimum difference between the values of any two nodes. Since the inorder traversal of a binary search tree is an increasing sequence, we only need to find the minimum difference between the values of two adjacent nodes in the inorder traversal.
We can use a recursive method to implement the inorder traversal. During the process, we use a variable $\textit{pre}$ to save the value of the previous node. This way, we can calculate the minimum difference between the values of two adjacent nodes during the traversal.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary search tree.
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21 # 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 : Optional [ TreeNode ]):
if root is None :
return
dfs ( root . left )
nonlocal pre , ans
ans = min ( ans , root . val - pre )
pre = root . val
dfs ( root . right )
pre = - inf
ans = inf
dfs ( root )
return ans
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35 /**
* 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 final int inf = 1 << 30 ;
private int ans = inf ;
private int pre = - inf ;
public int minDiffInBST ( TreeNode root ) {
dfs ( root );
return ans ;
}
private void dfs ( TreeNode root ) {
if ( root == null ) {
return ;
}
dfs ( root . left );
ans = Math . min ( ans , root . val - pre );
pre = root . val ;
dfs ( root . right );
}
}
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29 /**
* 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 :
int minDiffInBST ( TreeNode * root ) {
const int inf = 1 << 30 ;
int ans = inf , pre = - inf ;
auto dfs = [ & ]( auto && dfs , TreeNode * root ) -> void {
if ( ! root ) {
return ;
}
dfs ( dfs , root -> left );
ans = min ( ans , root -> val - pre );
pre = root -> val ;
dfs ( dfs , root -> right );
};
dfs ( dfs , root );
return ans ;
}
};
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24 /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func minDiffInBST ( root * TreeNode ) int {
const inf int = 1 << 30
ans , pre := inf , - inf
var dfs func ( * TreeNode )
dfs = func ( root * TreeNode ) {
if root == nil {
return
}
dfs ( root . Left )
ans = min ( ans , root . Val - pre )
pre = root . Val
dfs ( root . Right )
}
dfs ( root )
return ans
}
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28 /**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function minDiffInBST ( root : TreeNode | null ) : number {
let [ ans , pre ] = [ Infinity , - Infinity ];
const dfs = ( root : TreeNode | null ) => {
if ( ! root ) {
return ;
}
dfs ( root . left );
ans = Math . min ( ans , root . val - pre );
pre = root . val ;
dfs ( root . right );
};
dfs ( root );
return ans ;
}
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40 // 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 :: cell :: RefCell ;
use std :: rc :: Rc ;
impl Solution {
pub fn min_diff_in_bst ( root : Option < Rc < RefCell < TreeNode >>> ) -> i32 {
const inf : i32 = 1 << 30 ;
let mut ans = inf ;
let mut pre = - inf ;
fn dfs ( node : Option < Rc < RefCell < TreeNode >>> , ans : & mut i32 , pre : & mut i32 ) {
if let Some ( n ) = node {
let n = n . borrow ();
dfs ( n . left . clone (), ans , pre );
* ans = ( * ans ). min ( n . val - * pre );
* pre = n . val ;
dfs ( n . right . clone (), ans , pre );
}
}
dfs ( root , & mut ans , & mut pre );
ans
}
}
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26 /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var minDiffInBST = function ( root ) {
let [ ans , pre ] = [ Infinity , - Infinity ];
const dfs = root => {
if ( ! root ) {
return ;
}
dfs ( root . left );
ans = Math . min ( ans , root . val - pre );
pre = root . val ;
dfs ( root . right );
};
dfs ( root );
return ans ;
};
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