Question

530. 二叉搜索树的最小绝对差

English Version

题目描述

给你一个二叉搜索树的根节点 root ，返回 树中任意两不同节点值之间的最小差值 。

差值是一个正数，其数值等于两值之差的绝对值。

 

示例 1：

输入：root = [4,2,6,1,3]
输出：1

示例 2：

输入：root = [1,0,48,null,null,12,49]
输出：1

 

提示：

• 树中节点的数目范围是 [2, 104]
• 0 <= Node.val <= 105

 

注意：本题与 783 https://leetcode.cn/problems/minimum-distance-between-bst-nodes/ 相同

解法

方法一：中序遍历

中序遍历二叉搜索树，获取当前节点与上个节点差值的最小值即可。

# 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

/**
 * 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);
  }
}

/**
 * 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);
  }
};

/**
 * 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
}

// 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);
    }
  }
}