Two Sum IV - Input is a BST LT653

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

Idea 1. Similar to Two Sum LT1, use set to record the element while looping the BST and check if target - num exists.

Time complexity: O(n)

Space complexity: O(h) + O(n)

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

         if(record.contains(k - root.val)) {
             return true;
         }
         record.add(root.val);
         return findHelper(root.left, k, record)
             || findHelper(root.right, k, record);
     }
     public boolean findTarget(TreeNode root, int k) {
         return findHelper(root, k, new HashSet<Integer>());
     }
 }

Idea 2 BFS + Set, insted of DFS(preorder, inorder, postordal) traversal like idea 1

Time complexity: O(n)

Space complexity: O(n)

 class Solution {
     public boolean findTarget(TreeNode root, int k) {
         if(root == null) {
             return false;
         }
         Deque<TreeNode> queue = new ArrayDeque<>();
         Set<Integer> record = new HashSet<>();
         queue.add(root);

         while(!queue.isEmpty()) {
             TreeNode node = queue.pollFirst();
             if(record.contains(k - node.val)) {
                     return true;
             }
             record.add(node.val);
             if(node.left != null) {
                 queue.offerLast(node.left);
             }
             if(node.right != null) {
                 queue.offerLast(node.right);
             }
         }

         return false;
     }
 }

Idea 3. Store the ordered nums in array after inorder traversal, then two pointer scannning towards each other.

Time complexity: O(n)

Space complexity: O(h) + O(n)

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 class Solution {
     private void inorderWalk(TreeNode root, List<Integer> nums) {
         if(root == null) {
             return;
         }

         inorderWalk(root.left, nums);
         nums.add(root.val);
         inorderWalk(root.right, nums);
     }

     public boolean findTarget(TreeNode root, int k) {
         List<Integer> nums = new ArrayList<>();

         inorderWalk(root, nums);

         for(int left = 0, right = nums.size()-1; left < right; ) {
             int sum = nums.get(left) + nums.get(right);
             if(sum == k) {
                 return true;
             }
             else if(sum < k) {
                 ++left;
             }
             else {
                 --right;
             }
         }

         return false;
     }
 }

Idea 4. Inorder walk iterator  + reversed inorder walk iterator to avoid extra space.

巧点： 1. 怎么停止循环？利用bst有序不重复，left < right; 否则 iter.hasNext()

2. 只有需要时才移动iterator

3. next() -> 遍历到下一个数就返回

Time complexity: O(n)

Space complexity: O(h)

 import java.util.NoSuchElementException;
 import java.lang.UnsupportedOperationException;
 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */

 class InorderIterator implements Iterator<TreeNode> {
     private TreeNode curr;
     private Deque<TreeNode> stack;
     private boolean forward;

     public InorderIterator(TreeNode root, boolean forward) {
         this.curr = root;
         this.stack = new ArrayDeque<>();
         this.forward = forward;
     }

     public boolean hasNext() {
         return curr!= null || !stack.isEmpty();
     }

     public TreeNode next() {
         if(!hasNext()) {
             throw new NoSuchElementException("tree run out of elements");
         }
         while(curr != null) {
             stack.push(curr);
             if(forward) {
                 curr = curr.left;
             }
             else {
                 curr = curr.right;
             }
         }
         curr = stack.pop();
         TreeNode prev = curr;
         if(forward) {
             curr = curr.right;
         }
         else {
             curr = curr.left;
         }
         return prev;
     }

     public void remove() {
         throw new UnsupportedOperationException();
     }
 }

 class Solution {
     public boolean findTarget(TreeNode root, int k) {
         InorderIterator iter = new InorderIterator(root, true);
         InorderIterator reversedIter = new InorderIterator(root, false);

         for(int left = iter.next().val, right = reversedIter.next().val; left < right;) {
             if(left + right == k) {
                 return true;
             }
             else if(left + right < k) {
                 left = iter.next().val;
             }
             else {
                 right = reversedIter.next().val;
             }
         }

         return false;
     }
 }

 import java.util.NoSuchElementException;
 import java.lang.UnsupportedOperationException;
 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */

 class InorderIterator implements Iterator<TreeNode> {
     private TreeNode curr;
     private Deque<TreeNode> stack;
     private boolean forward;

     public InorderIterator(TreeNode root, boolean forward) {
         this.curr = root;
         this.stack = new ArrayDeque<>();
         this.forward = forward;
     }

     public boolean hasNext() {
         return curr!= null || !stack.isEmpty();
     }

     public TreeNode next() {
         // if(!hasNext()) {
         //     throw new NoSuchElementException("tree run out of elements");
         // }
         while(curr != null || !stack.isEmpty()) {
             if(curr != null) {
                 stack.push(curr);
                 if(forward) {
                     curr = curr.left;
                 }
                 else {
                     curr = curr.right;
                 }
             }
             else {
                 curr = stack.pop();
                 TreeNode prev = curr;
                 if(forward) {
                     curr = curr.right;
                 }
                 else {
                     curr = curr.left;
                 }
                 return prev;
             }
         }
         throw new NoSuchElementException("tree run out of elements");
     }

     public void remove() {
         throw new UnsupportedOperationException();
     }
 }

 class Solution {
     public boolean findTarget(TreeNode root, int k) {
         InorderIterator forwardIter = new InorderIterator(root, true);
         InorderIterator backwardIter = new InorderIterator(root, false);

         for(int left = forwardIter.next().val, right = backwardIter.next().val; left < right; ) {
             int sum = left + right;
             if(sum  ==  k) {
                 return true;
             }
             else if (sum < k) {
                 left = forwardIter.next().val;
             }
             else {
                 right = backwardIter.next().val;
             }

         }

         return false;
     }
 }