173. Binary Search Tree Iterator

四刷 05/2022

Runtime: 21 ms, faster than 55.36% of Java online submissions for Binary Search Tree Iterator.

Memory Usage: 51.3 MB, less than 69.64% of Java online submissions for Binary Search Tree Iterator.

/**

 * 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 BSTIterator {

    Stack<TreeNode> stack;

    

    public BSTIterator(TreeNode root) {

        // For each tree, always iterate the left sub tree, then root, then right sub tree

        this.stack = new Stack<>();

        while (root != null) {

            stack.push(root);

            root = root.left;

        }

    }

    

    public int next() {

        TreeNode curr = stack.pop();

        int val = curr.val;

        curr = curr.right;

        while (curr != null) {

            stack.push(curr);

            curr = curr.left;

        }

        return val;

    }

    

    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();

 */

三刷
3.72 %
class BSTIterator {

    Stack<TreeNode> stack;
    public BSTIterator(TreeNode root) {
        stack = new Stack();
        pushAll(root);
    }
   
    /** @return the next smallest number */
    public int next() {
        TreeNode result = stack.pop();
        pushAll(result.right);
        return result.val;
    }
   
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }
   
    private void pushAll(TreeNode curr) {
        while (curr != null) {
            stack.push(curr);
            curr = curr.left;
        }
    }
}


二刷
33.99 %
public class BSTIterator {
    Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        TreeNode curr = root;
        stack = new Stack<>();
        pushAll(curr);
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode curr = stack.pop();
        int result = curr.val;
     
        curr = curr.right;
        pushAll(curr);
        return result;
    }
    private void pushAll(TreeNode curr) {
        while (curr != null) {
            stack.push(curr);
            curr = curr.left;
        }
    }
}

一刷
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {
    private Stack<TreeNode> stack;
    private TreeNode curr;
    public BSTIterator(TreeNode root) {
        this.stack = new Stack<TreeNode>();
        this.curr = root;
    }

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

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

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