力扣面试150题--二叉搜索树迭代器

Day 50

题目描述

思路

初次思路：想的比较简单，在构造这个类的时候，直接求出中序遍历，存放在一个数组中，维护一个序号，当然这个不满足进阶做法的空间复杂度，因为需要保存中序遍历的所有值。

/**
 * 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 {
    public TreeNode root;
    public List<TreeNode>mid;
    int i=0;
    public BSTIterator(TreeNode root) {
        mid=new ArrayList<TreeNode>();
        findmin(root);
    }
    public void findmin(TreeNode root){
        if(root==null){
            return;
        }
        findmin(root.left);
        mid.add(root);
        findmin(root.right);
    }
    public int next() {
        int res= mid.get(i).val;
        i++;
        return res;
    }
    
    public boolean hasNext() {
        if(i>=mid.size()){
            return false;
        }
        else{
            return true;
        }
    }
}

/**
 * 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();
 */

进阶做法：（题解）
迭代做法，我们知道取得中序遍历可以通过栈来实现，那么就把中序遍历采取非递归写法，每次获取下一个节点，就从栈中取出一个节点，并且处理它后面需要压入栈的节点处理了。这样就满足了进阶的空间复杂度。

class BSTIterator {
    private TreeNode cur;
    private Deque<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        cur = root;
        stack = new LinkedList<TreeNode>();
    }
    
    public int next() {
        while (cur != null) {
            stack.push(cur);
            cur = cur.left;
        }
        cur = stack.pop();
        int ret = cur.val;
        cur = cur.right;
        return ret;
    }
    
    public boolean hasNext() {
        return cur != null || !stack.isEmpty();
    }
}