Binary Search Tree Iterator

问题描述

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.

 

解决思路

中序遍历+非递归（辅助栈）

时间/空间复杂度均为O(logn).

 

程序

public class BSTIterator {
    private Stack<TreeNode> s;
    
    public BSTIterator(TreeNode root) {
        s = new Stack<>();
        pushLeft(root);
    }
    
    private void pushLeft(TreeNode node) {
        TreeNode cur = node;
        while (cur != null) {
            s.push(cur);
            cur = cur.left;
        }
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode node = s.pop();
        pushLeft(node.right);
        return node.val;
    }
}