Search Range in Binary Search Tr

Search Range in Binary Search Tree

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今天是一道有关二叉树的题目，来自LintCode，难度为Medium。

题目如下

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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.

解题思路及代码见阅读原文

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解题思路

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首先，老规矩，看到二叉树，想到二叉树的三种遍历方式。

然后，这是是按升序排序，很明显是中序遍历。

其次，知道了采用中序遍历，下面就要考虑两个问题：

• 每个节点如何处理，从题目中很容易看出，当该节点的值在k1和k2之间时加入链表，否则跳过。

• 递归的终止条件：比较明显，即当节点的值比k1小或者比k2大时终止。即此时，不在遍历该节点的左右子树。

最后，我们来看代码。

代码如下

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Java版

/**
 * 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>();
        midOrder(list, root, k1, k2);
        return list;
    }
        
    private void midOrder(List<Integer> ret, TreeNode root, int k1, int k2) {
        if(null == root)
            return;
        boolean biggerK1 = root.val >= k1;
        boolean smallerK2 = root.val <= k2;
        if(biggerK1)
            midOrder(ret, root.left, k1, k2);
        if(biggerK1 && smallerK2)
            ret.add(root.val);
        if(smallerK2)
            midOrder(ret, root.right, k1, k2);
    }
}