[LeetCode ] Minimum Absolute Difference in BST

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/GYH0730/article/details/84288920

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).
Note: There are at least two nodes in this BST.

题意：找出二叉搜索树种任意两个节点最小的差的绝对值。

思路：中序遍历一遍，得到个递增数组，然后找相邻元素差的最小值。

C代码：

int e[100005],tot;
void inOrderTraverse(struct TreeNode* root)
{
    if(root != NULL) {
        inOrderTraverse(root -> left);
        e[tot++] = root -> val;
        inOrderTraverse(root -> right);
    }
}
int min(int a,int b)
{
    return a >= b ? b : a;
}
int getMinimumDifference(struct TreeNode* root) {
    tot = 0;
    inOrderTraverse(root);
    int i,res = 0x3f3f3f3f;
    for(i = 1; i < tot; i++) {
        res = min(e[i] - e[i - 1],res);
    }
    return res;
}