LeetCode Two Sum IV - Input is a BST

原题链接在这里：https://leetcode.com/problems/two-sum-iv-input-is-a-bst/description/

题目：

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

题解：

BST inorder traversal 得到ascending的list, 用two pointers 夹比找k.

Time Complexity: O(n), n 是node数.

Space: O(n), list size.

AC Java:

/**
 * 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) {
        List<Integer> list = new ArrayList<Integer>();
        inorder(root, list);
        int l = 0; 
        int r = list.size()-1;
        while(l < r){
            if(list.get(l) + list.get(r) == k){
                return true;
            }else if(list.get(l) + list.get(r) < k){
                l++;
            }else{
                r--;
            }
        }
        return false;
    }
    
    private void inorder(TreeNode root, List<Integer> list){
        if(root == null){
            return;
        }
        inorder(root.left, list);
        list.add(root.val);
        inorder(root.right, list);
    }
}

或者对每个TreeNode cur在BST中找k-cur.val.

Time Complexity: O(nlogn). 

Space: O(logn), stack space.

AC Java:

/**
 * 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) {
        return dfs(root, root, k);
    }
    
    private boolean dfs(TreeNode root, TreeNode cur, int k){
        if(cur == null){
            return false;
        }
        return search(root, cur, k-cur.val) || dfs(root, cur.left, k) || dfs(root, cur.right, k);
    }
    
    private boolean search(TreeNode root, TreeNode cur, int target){
        if(root == null){
            return false;
        }
        
        if(root.val == target && root != cur){
            return true;
        }else if(root.val < target){
            return search(root.right, cur, target);
        }else if(root.val > target){
            return search(root.left, cur, target);
        }
        
        return false;
    }
}