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783. Minimum Distance Between BST Nodes

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

• Java
• C++
• Python
• Go
• Javascript

• /**
   * 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);
      }
  }
  

• /**
   * 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);
      }
  };
  

• # 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
  
  

• /**
   * 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
  }
  

• 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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