173.Binary search tree iterator

Posted 2020-09-29

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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

对中序遍历的操作:

/**
 * 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 = new Stack<TreeNode>();
    
    public BSTIterator(TreeNode root) {
        cur = root;
    }

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

    }

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

注意: return (cur != null || !stack.isEmpty());
root 先不push进去,为了next()方便