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.
思路:
遍历二叉树,将元素值排序,最小的差肯定出现在相邻两个元素里,遍历数组即可。
1 void preorderTree(TreeNode* root,vector<int>& vc) 2 { 3 if(root ==NULL) return ; 4 vc.push_back(root->val); 5 preorderTree(root->left,vc); 6 preorderTree(root->right,vc); 7 } 8 int minDiffInBST(TreeNode* root) 9 { 10 vector<int>vc; 11 preorderTree(root,vc); 12 sort(vc.begin(),vc.end()); 13 int t =INT_MAX; 14 for(int i=0;i<vc.size()-1;i++) 15 { 16 t = min(t,vc[i+1] - vc[i]); 17 } 18 return t; 19 }