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
思路：把值放入map中，遍历值，并判断另一个组合数是否在map中即可。
JAVA CODE
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
Map<Integer,Integer> map = new HashMap<>();
int key = 0;
public boolean findTarget(TreeNode root, int k) {
// Queue<TreeNode> queue = new ArrayDeque<>();
// queue.offer(root);
// while(!queue.isEmpty()){
// root =
// if()
// }
find(root);
for (Integer value : map.values()) {
if(value*2!=k&&map.containsValue(k - value.intValue()))
return true;
}
return false;
}
void find(TreeNode rr){
if(rr != null){
find(rr.left);
map.put(new Integer(key++), new Integer(rr.val));
find(rr.right);
}
}
}