Search Range in Binary Search Tree

Question

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

If k1 = 10 and k2 = 22, then your function should return [12, 20, 22].

  20
 /  \
8   22

/
4 12

题解 - 中序遍历

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。

C++

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/**
* 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);
}
};

Java

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/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @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.
*/
public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
ArrayList<Integer> result = new ArrayList<Integer>();
helper(root, k1, k2, result);

return result;
}

private void helper(TreeNode root, int k1, int k2, ArrayList<Integer> result) {
if (root == null) return;

// in-order binary tree iteration
helper(root.left, k1, k2, result);
if (k1 <= root.val && root.val <= k2) {
result.add(root.val);
}
helper(root.right, k1, k2, result);
}
}

源码分析

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:

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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
}

Warning 「剪枝」的判断条件容易出错,应将当前节点的值与k1k2进行比较而不是其左子节点或右子节点的值。




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