173. Binary Search Tree Iterator - Medium

发布时间：2020-12-14 05:17:37 所属栏目：大数据来源：网络整理

导读：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. ? Example: BSTIterator iterator = new BSTIterator(root)

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.

Example:

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.

M1:?先inorder全部遍历完（慢）

time: O(n),?space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class BSTIterator {
    List<Integer> list;
    int idx;

    public BSTIterator(TreeNode root) {
        list = new ArrayList<>();
        idx = 0;
        inorder(root,list);
    }
    
    public void inorder(TreeNode root,List<Integer> list) {
        if(root == null) {
            return;
        }
        inorder(root.left,list);
        list.add(root.val);
        inorder(root.right,list);
    }
    
    /** @return the next smallest number */
    public int next() {
        int tmp = idx;
        idx++;
        return list.get(tmp);
    }
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return idx < list.size();
    }
}

/**
 * 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();
 */

M2: optimized，用stack，先把所有的左边的节点压进栈，当call next()时再按顺序出栈

next()　　　? ?-- time: O(h),space: O(h)

hasNext()　　-- time: O(1)

/**
 * 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 = new Stack<>();

    public BSTIterator(TreeNode root) {
        getAllLeft(root);
    }
    
    public void getAllLeft(TreeNode root) {
        while(root != null) {
            stack.push(root);
            root = root.left;
        }
    }
    
    /** @return the next smallest number */
    public int next() {
        TreeNode node = stack.pop();
        getAllLeft(node.right);       
        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();
 */

（编辑：李大同）

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