Binary Search Tree Iterator

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

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

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

分析

就是inorder的while遍历拆成2部分。

public class BSTIterator {
    private TreeNode next;
    private Stack<TreeNode> s = new Stack<TreeNode>();

    private void addToStack(){
        while(next != null){
            s.add(next);
            next = next.left;
        }
        next = null;
    }

    public BSTIterator(TreeNode root) {
        next = root;
    }

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

    /** @return the next smallest number */
    public int next() {
        if(!hasNext()){
            return -1;
        }
        next = s.pop();
        int ret = next.val;
        next = next.right;
        return ret;

    }
}

不用额外函数的做法

二刷错误：1.忘了用栈 2. hasNext里的while写错了，root不空的话，先推入栈再转左。

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

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

    /** @return whether we have a next smallest number */
    public boolean hasNext() {        
        while(cur != null){
            s.push(cur);
            cur = cur.left;
        }

        return !s.isEmpty();
    }

    /** @return the next smallest number */
    public int next() {
        if(!s.isEmpty()){
            cur = s.pop(); 
            int ret = cur.val;
            cur = cur.right;
            return ret;
        }
        return -1;
    }
}

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