173. Binary Search Tree Iterator.md

173. Binary Search Tree Iterator

Question

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

Thinking:

• Method:
  • 使用LinkedList装载中序遍历的结果。O(N) extra space

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {
    private LinkedList<Integer> list;
    private ListIterator<Integer> cur = null;
    public BSTIterator(TreeNode root) {
        list = new LinkedList<>();
        inorder(root, list);
        cur = list.listIterator(0);
    }
    private void inorder(TreeNode node, List<Integer> result){
        if(node == null) return;
        inorder(node.left, result);
        result.add(node.val);
        inorder(node.right, result);
    }
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return cur.hasNext();
    }

    /** @return the next smallest number */
    public int next() {
        return cur.next();
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */

• Method 2： 通过栈实现

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {
    Stack<TreeNode> stack;
    public BSTIterator(TreeNode root) {
        stack = new Stack<>();
        while(root != null){
            stack.push(root);
            root = root.left;
        }
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode cur = stack.pop();
        if(cur.right != null){
            TreeNode next = cur.right;
            while(next != null){
                stack.push(next);
                next = next.left;
            }
        }
        return cur.val;
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */

二刷

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class BSTIterator {
    private List<Integer> inorder = new LinkedList<>();
    private ListIterator<Integer> cur = null;
    public BSTIterator(TreeNode root) {
        inorder(root, inorder);
        cur = inorder.listIterator(0);
    }
    private void inorder(TreeNode root, List<Integer> list){
        if(root == null) return;
        inorder(root.left, list);
        list.add(root.val);
        inorder(root.right, list);
    }
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return cur.hasNext();
    }
    /** @return the next smallest number */
    public int next() {
        return cur.next();
    }
}
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */

Third Time

• Method 1: Recursion

   /**
    * Definition for a binary tree node.
    * public class TreeNode {
    *     int val;
    *     TreeNode left;
    *     TreeNode right;
    *     TreeNode(int x) { val = x; }
    * }
    */
   class BSTIterator {
       private Queue<Integer> q;
       public BSTIterator(TreeNode root) {
           q = new LinkedList<>();
           inorder(root);
       }
       private void inorder(TreeNode node){
           if(node == null) return;
           inorder(node.left);
           q.offer(node.val);
           inorder(node.right);
       }
  
       /** @return the next smallest number */
       public int next() {
           return q.poll();
       }
  
       /** @return whether we have a next smallest number */
       public boolean hasNext() {
           return !q.isEmpty();
       }
   }
  
   /**
    * Your BSTIterator object will be instantiated and called as such:
    * BSTIterator obj = new BSTIterator(root);
    * int param_1 = obj.next();
    * boolean param_2 = obj.hasNext();
    */

• Method 2: loop

   /**
    * Definition for a binary tree node.
    * public class TreeNode {
    *     int val;
    *     TreeNode left;
    *     TreeNode right;
    *     TreeNode(int x) { val = x; }
    * }
    */
   class BSTIterator {
       Stack<TreeNode> stack;
       public BSTIterator(TreeNode root) {
           this.stack = new Stack<>();
           if(root == null) return;
           stack.push(root);
           while(root.left != null){
               stack.push(root.left);
               root = root.left;
           }
       }
  
       /** @return the next smallest number */
       public int next() {
           TreeNode node = stack.pop();
           if(node.right != null){
               TreeNode temp = node.right;
               stack.push(temp);
               while(temp.left != null){
                   stack.push(temp.left);
                   temp = temp.left;
               }
           }
           return node.val;
       }
  
       /** @return whether we have a next smallest number */
       public boolean hasNext() {
           return !stack.isEmpty();
       }
   }
  
   /**
    * Your BSTIterator object will be instantiated and called as such:
    * BSTIterator obj = new BSTIterator(root);
    * int param_1 = obj.next();
    * boolean param_2 = obj.hasNext();
    */