Given a Binary Search Tree (BST) with the root node root, return the minimum difference between the values of any two different nodes in the tree.
Example :
Input: root = [4,2,6,1,3,null,null]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array. The given tree [4,2,6,1,3,null,null] is represented by the following diagram: 4
/ \
2 6
/ \
1 3 while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.
Note:
- The size of the BST will be between 2 and
100. - The BST is always valid, each node's value is an integer, and each node's value is different.
这道题跟之前那道Minimum Absolute Difference in BST没有任何区别，解法完全可以共用，讲解也可以参见之前的帖子，这里就简略的说一下。第一种方法很直接，通过中序遍历按顺序从小到大将所有的结点值都存入到一个数组中，然后就遍历这个数组，找相邻的两个的差值最小的返回即可，参见代码如下：
解法一：
class Solution {
public:
int minDiffInBST(TreeNode* root) {
int res = INT_MAX;
vector<int> v;
helper(root, v);
for (int i = ; i < v.size(); ++i) {
res = min(res, v[i] - v[i - ]);
}
return res;
}
void helper(TreeNode* node, vector<int>& vals) {
if (!node) return;
helper(node->left, vals);
vals.push_back(node->val);
helper(node->right, vals);
}
};
我们可以优化上面解法的空间复杂度，并不记录所有的结点值，而是只记录之前的结点值，然后做差值更新结果res即可。
解法二：
class Solution {
public:
int minDiffInBST(TreeNode* root) {
int res = INT_MAX, pre = -;
helper(root, pre, res);
return res;
}
void helper(TreeNode* node, int& pre, int& res) {
if (!node) return;
helper(node->left, pre, res);
if (pre != -) res = min(res, node->val - pre);
pre = node->val;
helper(node->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：
解法三：
class Solution {
public:
int minDiffInBST(TreeNode* root) {
int res = INT_MAX;
helper(root, INT_MIN, INT_MAX, res);
return res;
}
void helper(TreeNode* node, int low, int high, int& res) {
if (!node) return;
if (low != INT_MIN) res = min(res, node->val - low);
if (high != INT_MAX) res = min(res, high - node->val);
helper(node->left, low, node->val, res);
helper(node->right, node->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：
解法四：
class Solution {
public:
int minDiffInBST(TreeNode* root) {
int res = INT_MAX, pre = -;
stack<TreeNode*> st;
TreeNode* p = root;
while (!st.empty() || p) {
if (p) {
st.push(p);
p = p->left;
} else {
p = st.top(); st.pop();
if (pre != -) res = min(res, p->val - pre);
pre = p->val;
p = p->right;
}
}
return res;
}
};
类似题目：
Minimum Absolute Difference in BST
参考资料：
https://leetcode.com/problems/minimum-distance-between-bst-nodes/solution/