Minimum Absolute Difference in BST问题及解法

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问题描述：

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

示例：

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

问题分析：

BST本身就排好序了，最小差值只会在排好序的序列中相邻两个值中出现。

过程详见代码：

/*** 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 INT_MAX;TreeNode *right = findRight(root->left);TreeNode *left = findLeft(root->right);if(right != NULL && left != NULL){int m1 = min(abs(root->val - right->val),abs(root->val - left->val));int m2 = min(getMinimumDifference(root->left),getMinimumDifference(root->right));return min(m1,m2); }else if(right != NULL){int m1 = abs(root->val - right->val);int m2 = getMinimumDifference(root->left);return min(m1,m2);}else if(left != NULL){int m1 = abs(root->val - left->val);int m2 = getMinimumDifference(root->right);return min(m1,m2);}else{return INT_MAX;}}TreeNode * findRight(TreeNode* root){if(root == NULL) return NULL;while(root->right != NULL) root = root->right;return root; }TreeNode * findLeft(TreeNode* root){if(root == NULL) return NULL;while(root->left != NULL) root = root->left;return root; }
};

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