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;
}
}