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.
/**
* 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) {
// write your code here
vector<int> seen;
search(root, k1, k2, seen);
return seen;
}
void search(TreeNode* root, int k1, int k2, vector<int>& seen) {
if (k2 < k1 || root == NULL) {
return;
}
if (k1 >= root->val) {
// range all in right tree
if (k1 == root->val) {
seen.push_back(k1);
}
search(root->right, k1, k2, seen);
} else if (k2 <= root->val) {
// range all in left tree
search(root->left, k1, k2, seen);
if (k2 == root->val) {
seen.push_back(k2);
}
} else {
// normal
search(root->left, k1, root->val, seen);
seen.push_back(root->val);
search(root->right, root->val, k2, seen);
}
}
};
把情况合并一下可以写成如下形式:
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) {
// write your code here
vector<int> seen;
search(root, k1, k2, seen);
return seen;
}
void search(TreeNode* root, int k1, int k2, vector<int>& seen) {
if (k2 < k1 || root == NULL) {
return;
}
if (k1 < root->val) {
search(root->left, k1, min(root->val, k2), seen);
}
if (k1 <= root->val && root->val <= k2) {
seen.push_back(root->val);
}
if (k2 > root->val) {
search(root->right, max(root->val, k1), k2, seen);
}
}
};