Two Sum IV - Input is a BST

题目描述：

Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.

Example 1:

Input: 
    5
   / \
  3   6
 / \   \
2   4   7

Target = 9

Output: True

Example 2:

Input: 
    5
   / \
  3   6
 / \   \
2   4   7

Target = 28

Output: False

分析：

直接在二叉树上进行搜索比较困难，可以先采用前序遍历将二叉树转换成有序数组，然后采用首位指针的方式搜索结果。

代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
     public boolean findTarget(TreeNode root, int k) {
        if(root == null){
            return false;
        }

        List<Integer> result = BST2Array(root);
        if(result == null || result.size() <= 1){
            return false;
        }
        // 首尾遍历有序数组
        int i = 0;
        int j = result.size()-1;
        while (i < j){
            int sum = result.get(i) + result.get(j);
            if(sum == k){
                return true;
            }
            if(sum < k){
                i++;
            }else {
                j--;
            }

        }
        return false;
    }

    private  List<Integer> BST2Array(TreeNode root) {
        if(root == null) {
            return null;
        }
        // 前序遍历
        List<Integer> result = new ArrayList<>();
        List<Integer> left = BST2Array(root.left);
        List<Integer> right = BST2Array(root.right);

        if(left != null){
            result.addAll(left);
        }
        result.add(root.val);

        if(right!= null){
            result.addAll(right);
        }
        return result;
    }
}