Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search Tree.
Find all the keys of tree in range k1 to k2. i.e. print all x such that k1<=x<=k2 and x is a key of given BST.
Return all the keys in ascending order.
Example
For example, if k1 = 10 and k2 = 22, then your function should print 12, 20 and 22.
20
/ \
8 22
/ \
4 12
中等偏易难度题,本题涉及到二叉查找树的按序输出,应马上联想到二叉树的中序遍历,对于二叉查找树而言,使用中序遍历即可得到有序元素。对每次访问的元素加以判断即可得最后结果,由于 OJ 上给的模板不适合递归处理,新建一个私有方法即可。
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param root: The root of the binary search tree.
* @param k1 and k2: range k1 to k2.
* @return: Return all keys that k1<=key<=k2 in ascending order.
*/
vector<int> searchRange(TreeNode* root, int k1, int k2) {
vector<int> result;
inorder_dfs(result, root, k1, k2);
return result;
}
private:
void inorder_dfs(vector<int> &ret, TreeNode *root, int k1, int k2) {
if (NULL == root) {
return;
}
inorder_dfs(ret, root->left, k1, k2);
if ((root->val >= k1) && (root->val <= k2)) {
ret.push_back(root->val);
}
inorder_dfs(ret, root->right, k1, k2);
}
};
以上为题解思路的简易实现,可以优化的地方为「剪枝过程」的处理——不递归遍历不可能有解的节点。优化后的inorder_dfs
如下:
void inorder_dfs(vector<int> &ret, TreeNode *root, int k1, int k2) {
if (NULL == root) {
return;
}
if ((NULL != root->left) && (root->val > k1)) {
inorder_dfs(ret, root->left, k1, k2);
} // cut-off for left sub tree
if ((root->val >= k1) && (root->val <= k2)) {
ret.push_back(root->val);
} // add valid value
if ((NULL != root->right) && (root->val < k2)) {
inorder_dfs(ret, root->right, k1, k2);
} // cut-off for right sub tree
}
「剪枝」的判断条件容易出错,应将当前节点的值与
k1
和k2
进行比较而不是其左子节点或右子节点的值。