資訊專欄INFORMATION COLUMN
摘要：重點(diǎn)是根據(jù)的性質(zhì)，先左后根最后右。另一重點(diǎn)是，函數(shù)和函數(shù)都要用的的參數(shù)，記得在函數(shù)外層定義。
Problem
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.
ExampleIf k1 = 10 and k2 = 22, then your function should return [12, 20, 22].
20 / 8 22 / 4 12Note
重點(diǎn)是：根據(jù)BST的性質(zhì)，先左后根最后右。
另一重點(diǎn)是，helper函數(shù)和main函數(shù)都要用的的參數(shù)，記得在main函數(shù)外層定義。
public class Solution {
private ArrayList result;
public ArrayList searchRange(TreeNode root, int k1, int k2) {
// write your code here
result = new ArrayList();
Searchfunc(root, k1, k2);
return result;
}
public void Searchfunc(TreeNode root, int k1, int k2) {
if (root == null) {
return;
}
if (root.val > k1) {
Searchfunc(root.left, k1, k2);
}
if (root.val >= k1 && root.val <= k2) {
result.add(root.val);
}
if (root.val < k2) {
Searchfunc(root.right, k1, k2);
}
}
}
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