LeetCode 173. Binary Search Tree Iterator

Question

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

Explanation

完成中序遍历，用stack这个数据结构存储。

Code

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

public class BSTIterator {
    Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
         stack = new Stack<>();
         pushAllLeft(root);
    }

    public void pushAllLeft(TreeNode node) {
        while(node != null) {
            stack.push(node);
            node = node.left;
        }
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        if(stack.isEmpty()) return false;
        return true;

    }

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

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