[Algo Rhythm🕺💃] LeetCode 653. Two Sum IV - Input is a BST

📚 Problem description

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

 

🎯 Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean findTarget(TreeNode root, int k) {
        List<Integer> list = new ArrayList<>();
        inorder(root, list);
        int left = 0, right = list.size() - 1;
        
        while(left < right) {
            int sum = list.get(left) + list.get(right);
            if(sum == k) return true;
            if(sum > k) right--;
            if(sum < k) left++;
        }
        
        return false;
    }
    public void inorder(TreeNode root, List<Integer> list) {
        if(root == null) return;
        inorder(root.left, list);
        list.add(root.val);
        inorder(root.right, list);
    }
}

 

🧐 Approach

BST is half-sorted data structure. We can get sorted array from BST by in-order traversal.

After getting sorted array, we can use two pointer approach to find two indexes.

 

📖  What I learned