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 :
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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: In order traversal
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua // Running time: 783 class Solution { public: int minDiffInBST(TreeNode* root) { min_diff_ = INT_MAX; prev_ = nullptr; minDiff(root); return min_diff_; } private: int min_diff_; int* prev_; void minDiff(TreeNode* root) { if (root == nullptr) return; minDiff(root->left); if (prev_ != nullptr) min_diff_ = min(min_diff_, abs(root->val - *prev_)); prev_ = &root->val; minDiff(root->right); } }; |
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