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.
Have you met this question in a real interview?
Yes
Example
If k1 = 10 and k2 = 22, then your function should return[12, 20, 22].
20
/ \
8 22
/ \
4 12
Tags
分析：
这题还是很简单的，只要根据root的值和k1, k2的值分三种情况讨论就可以了。
/**
* 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) {
// write your code here
ArrayList<Integer> list = new ArrayList<Integer>();
if (k1 <= k2) {
traverse(root, k1, k2, list);
}
Collections.sort(list);
return list;
} public void traverse(TreeNode node, int k1, int k2, ArrayList<Integer> list) {
if (node != null) {
if (node.val < k1) {
traverse(node.right, k1, k2, list);
} else if (node.val > k2) {
traverse(node.left, k1, k2, list);
} else {
list.add(node.val);
traverse(node.left, k1, k2, list);
traverse(node.right, k1, k2, list);
}
}
}
}
参考请注明出处：cnblogs.com/beiyeqingteng/