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

我的做法是inorder traversal的变形,判断是否向左边递归的时候加上判断是否:root.val > k1, 如果否,则不需要继续向左递归;右子树的处理方法类似

第一次做法,把result数组作为return type,不好,消耗额外空间

 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> res = searchRangeRecur(root,k1,k2);
return res;
} public ArrayList<Integer> searchRangeRecur(TreeNode cur, int k1, int k2){
ArrayList<Integer> res = new ArrayList<Integer>();
if (cur==null) return res;
if (k1>k2) return res; ArrayList<Integer> left = searchRangeRecur(cur.left,k1,Math.min(cur.val-1,k2));
ArrayList<Integer> right = searchRangeRecur(cur.right,Math.max(cur.val+1,k1),k2); res.addAll(left);
if (cur.val>=k1 && cur.val<=k2) res.add(cur.val);
res.addAll(right); return res;
} }

第二遍做法:

 /**
* 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> res = new ArrayList<Integer>();
if (root == null || (k1 > k2)) return res;
helper(root, k1, k2, res);
return res;
} public void helper(TreeNode root, int k1, int k2, ArrayList<Integer> res) {
if (root == null) return;
helper(root.left, k1, k2, res);
if (k1 <= root.val && root.val <= k2) res.add(root.val);
helper(root.right, k1, k2, res);
}
}

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