[leetcode] Binary Search Tree

题目：（Tree Stack）

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.

题解：

tree的题很容易就和stack联系在一起了，这道题就是维护一个stack就可以。

用pushleft()

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

public class BSTIterator {
    //TreeNode root;
    Stack<TreeNode> stack = new Stack<TreeNode>();
    public BSTIterator(TreeNode root) {
        pushLeft(root);
    }

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

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

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

不断地将root.left push 到stack，当回溯的时候如果当时那个node有right child，那么多那个right child 再进行pushleft()。