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Formatted question description: https://leetcode.ca/all/783.html
783. Minimum Distance Between BST Nodes (Medium)
Given a Binary Search Tree (BST) with the root node root
, return the minimum difference between the values of any two different nodes in the tree.
Example :
Input: root = [4,2,6,1,3,null,null]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array.
The given tree [4,2,6,1,3,null,null] is represented by the following diagram:
4
/ \
2 6
/ \
1 3
while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.
Note:
- The size of the BST will be between 2 and
100
. - The BST is always valid, each node’s value is an integer, and each node’s value is different.
Solution 1.
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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int minDiffInBST(TreeNode root) { List<Integer> inorderTraversal = inorderTraversal(root); int minDiff = Integer.MAX_VALUE; int size = inorderTraversal.size(); for (int i = 1; i < size; i++) { int diff = inorderTraversal.get(i) - inorderTraversal.get(i - 1); minDiff = Math.min(minDiff, diff); } return minDiff; } public List<Integer> inorderTraversal(TreeNode root) { List<Integer> inorderTraversal = new ArrayList<Integer>(); Stack<TreeNode> stack = new Stack<TreeNode>(); TreeNode node = root; while (!stack.isEmpty() || node != null) { while (node != null) { stack.push(node); node = node.left; } TreeNode visitNode = stack.pop(); inorderTraversal.add(visitNode.val); node = visitNode.right; } return inorderTraversal; } } ############ /** * 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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// OJ: https://leetcode.com/problems/minimum-distance-between-bst-nodes/ // Time: O(N) // Space: O(log(N)) class Solution { private: int ans = INT_MAX; TreeNode *prev = NULL; void inorder(TreeNode *root) { if (!root) return; inorder(root->left); if (prev) ans = min(ans, root->val - prev->val); prev = root; inorder(root->right); } public: int minDiffInBST(TreeNode* root) { inorder(root); return ans; } };
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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 ############ # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def minDiffInBST(self, root): """ :type root: TreeNode :rtype: int """ vals = [] def inOrder(root): if not root: return inOrder(root.left) vals.append(root.val) inOrder(root.right) inOrder(root) return min([vals[i + 1] - vals[i] for i in xrange(len(vals) - 1)])
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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 min(a, b int) int { if a < b { return a } return b } func abs(x int) int { if x < 0 { return -x } return x }