530. Minimum Absolute Difference in BST

 

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.

 

这道题给了我们一棵二叉搜索树,让我们求任意个节点值之间的最小绝对差。由于BST的左<根<右的性质可知,如果按照中序遍历会得到一个有序数组,那么最小绝对差肯定在相邻的两个节点值之间产生。所以我们的做法就是对BST进行中序遍历,然后当前节点值和之前节点值求绝对差并更新结果res。这里需要注意的就是在处理第一个节点值时,由于其没有前节点,所以不能求绝对差。这里我们用变量pre来表示前节点值,这里由于题目中说明了所以节点值不为负数,所以我们给pre初始化-1,这样我们就知道pre是否存在。如果没有题目中的这个非负条件,那么就不能用int变量来,必须要用指针,通过来判断是否为指向空来判断前结点是否存在。还好这里简化了问题,用-1就能搞定了,这里我们先来看中序遍历的递归写法,参见代码如下:

 

解法一:

class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        int res = INT_MAX, pre = -1;
        inorder(root, pre, res);
        return res;
    }
    void inorder(TreeNode* root, int& pre, int& res) {
        if (!root) return;
        inorder(root->left, pre, res);
        if (pre != -1) res = min(res, root->val - pre);
        pre = root->val;
        inorder(root->right, pre, res);
    }
};

 

其实我们也不必非要用中序遍历不可,用先序遍历同样可以利用到BST的性质,我们带两个变量low和high来分别表示上下界,初始化为int的极值,然后我们在递归函数中,分别用上下界和当前节点值的绝对差来更新结果res,参见代码如下:

 

解法二:

class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        int res = INT_MAX;
        helper(root, INT_MIN, INT_MAX, res);
        return res;
    }
    void helper(TreeNode* root, int low, int high, int& res) {
        if (!root) return;
        if (low != INT_MIN) res = min(res, root->val - low);
        if (high != INT_MAX) res = min(res, high - root->val);
        helper(root->left, low, root->val, res);
        helper(root->right, root->val, high, res);
    }
};

 

下面这种方法是解法一的迭代的写法,思路跟之前的解法没有什么区别,参见代码如下:

 

解法三:

class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        int res = INT_MAX, pre = -1;
        stack<TreeNode*> st;
        TreeNode *p = root;
        while (p || !st.empty()) {
            while (p) {
                st.push(p);
                p = p->left;
            }
            p = st.top(); st.pop();
            if (pre != -1) res = min(res, p->val - pre);
            pre = p->val;
            p = p->right;
        }
        return res;
    }
};

 

参考资料:

https://discuss.leetcode.com/topic/80896/my-solution-using-no-recursive-in-order-binary-tree-iteration

https://discuss.leetcode.com/topic/80823/two-solutions-in-order-traversal-and-a-more-general-way-using-treeset/2

https://discuss.leetcode.com/topic/80916/java-no-in-order-traverse-solution-just-pass-upper-bound-and-lower-bound

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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