Minimum Absolute Difference in BST问题及解法

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
/
2

Output:
1

Explanation:
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;
}
};