leetcode-Minimum Absolute Difference in BST

Question:
Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.

Example:

Input:

1
\
3
/
2

Output:
1

Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).

Solution:

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int getMinimumDifference(TreeNode* root) {        if(root == NULL){            return 0x7fffffff;        }        int d = 0x7fffffff;        int lmin = 0x7fffffff;        int rmin = 0x7fffffff;        if(root->left){            TreeNode* pre = root->left;            TreeNode* cur = pre->right;            while(cur){                pre = cur;                cur = cur->right;            }            lmin = root->val - pre->val;        }        if(root->right){            TreeNode* pre = root->right;            TreeNode* cur = pre->left;            while(cur){                pre = cur;                cur = cur->left;            }            rmin = pre->val - root->val;        }        d = lmin<rmin?lmin:rmin;        int d1 = getMinimumDifference(root->left);        int d2 = getMinimumDifference(root->right);        int tmp = d1 < d2 ? d1:d2;        return tmp < d?tmp:d;    }};

总结：

虽然该题定位为easy，不过我认为还是有一定难度的，主要考察二叉搜索树，及其递归函数的用法。解可分为三部分，包含该节点的最小距离，该节点左子树的最小距离，该节点右子树距离，然后取这三个值最小值。包含该节点的最小距离 d =min{该节点 - 左子树中的最右下节点，右子树最左下节点 - 该节点}，然后左右子树递归分别得到d1 d2，最后再求最小值min{d,d1,d2}。