Binary Search Tree Iterator

Source

• lintcode: (86) Binary Search Tree Iterator

Design an iterator over a binary search tree with the following rules:

- Elements are visited in ascending order (i.e. an in-order traversal)
- next() and hasNext() queries run in O(1) time in average.

Example
For the following binary search tree, in-order traversal by using iterator is [1, 6, 10, 11, 12]

           10
         /    \
        1      11
         \       \
             6       12

Challenge
Extra memory usage O(h), h is the height of the tree.

Super Star: Extra memory usage O(1)

题解 - 中序遍历

仍然考的是中序遍历，但是是非递归实现。其实这道题等价于写一个二叉树中序遍历的迭代器。需要内置一个栈，一开始先存储到最左叶子节点的路径。在遍历的过程中，只要当前节点存在右子树，则进入右子树，存储从此处开始到当前子树里最左叶子节点的路径。

Java

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Example of iterate a tree:
 * Solution iterator = new Solution(root);
 * while (iterator.hasNext()) {
 *    TreeNode node = iterator.next();
 *    do something for node
 * } 
 */
public class Solution {
    private Stack<TreeNode> stack = new Stack<>();
    private TreeNode curt;

    // @param root: The root of binary tree.
    public Solution(TreeNode root) {
        curt = root;
    }

    //@return: True if there has next node, or false
    public boolean hasNext() {
        return (curt != null || !stack.isEmpty()); //important to judge curt != null
    }

    //@return: return next node
    public TreeNode next() {
        while (curt != null) {
            stack.push(curt);
            curt = curt.left;
        }

        curt = stack.pop();
        TreeNode node = curt;
        curt = curt.right;

        return node;
    }
}

源码分析

1. 这里容易出错的是 hasNext() 函数中的判断语句，不能漏掉 curt != null。
2. 如果是 leetcode 上的原题，由于接口不同，则不需要维护 current 指针。