【小熊刷題】Binary Search Tree Iteration

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.

*Difficulty: Medium
https://leetcode.com/problems/binary-search-tree-iterator/

Solution

被O(1)矇騙了，它說的是averageO(1),以爲不可以dfs之後的child，不過一開始的思路是對的，用stack，把左邊最小的放進stack去，看next的時候pop出來，然後再把right child及其左邊的node從大到小放進去。因爲stack是FILO的原則。

上code：

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

public class BSTIterator {
    public Stack<TreeNode> stack = new Stack<TreeNode>();
    public BSTIterator(TreeNode root) {
        stack = new Stack<TreeNode>();
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
    }

    public boolean hasNext() {
        return !stack.isEmpty();
    }

    public int next() {
        TreeNode node = stack.pop();
        int r = node.val;
        if(node.right != null){
            node = node.right;
            while(node != null){
                stack.push(node);
                node = node.left;
            } 
        }

        return r;
    }
}

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