173. Binary Search Tree Iterator

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Problem

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.

BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false

Note:

next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree. You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

Code

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

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

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

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

        cur = stack.pop();
        int val = cur.val;
        cur = cur.right;
        return val;
    }
}

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