Search Range in Binary Search Tree

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.

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Example

If k1 = 10 and k2 = 22, then your function should return[12, 20, 22].

    20
   /  \
  8   22
 / \
4   12

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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/