leetcode 173-Binary Search Tree Iterator(medium)

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.

 

这是从leetcode上看到的别人的解法，觉得很厉害，po在这里。https://leetcode.com/problems/binary-search-tree-iterator/discuss/52525/My-solutions-in-3-languages-with-Stack

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

public class BSTIterator {
    Stack<TreeNode> stack=new Stack<>();

    public BSTIterator(TreeNode root) {
        pushAll(root);
    }
    public void pushAll(TreeNode root) {
        while(root!=null){
            stack.push(root);
            root=root.left;
        }
    }

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

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

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