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.
- This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/
A:
第一遍,我是用vector save 再对比的
第二遍, 省掉了O(n) space. 只用2个变量代表前一个数和最终结果。 (同时我们不能假设preValue 是几,因此不能用其做flag.来表示是否已经找到left most. 当然更不能用minDiff. 因为就要改minDiff的值。因此多引入了一个变量findLeftMost )
/** * 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) { int minDiff = INT_MAX; int preVal = INT_MAX; bool findLeftMost = false; // use extra flag, in case left most is the INT_MIN helper(root, preVal, minDiff, findLeftMost); return minDiff; } private: void helper(TreeNode* root, int & preVal, int & minDiff, bool & findLeftMost) { if(root==nullptr) return; helper(root->left, preVal, minDiff,findLeftMost); if(findLeftMost ) // if not the left-most node { minDiff = min(minDiff, root->val - preVal); } findLeftMost = true; preVal = root->val; helper(root->right, preVal, minDiff, findLeftMost); } };
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