leetcode-173：Binary Search Tree Iterator（Java）

Binary Search Tree Iterator

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.

Credits:

Special thanks to @ts for adding this problem and creating all test cases.

要求：写一个二叉查找树，每次返回树中的下一个最小节点

比如上图中的二叉查找树，从根节点开始，依次返回1,3,4,6,7... ...

思路：维护一个栈，先将根结点的左子树全部压栈，每次弹出栈顶元素，若某次弹出的栈顶元素有右子树，比如3，此时需要将以该节点的右子树为根的子树的左子节点全部压栈

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

import java.util.Stack;
public class BSTIterator {
    Stack<TreeNode> stack = new Stack<TreeNode>();

    public BSTIterator(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 minCurrent = stack.pop();
        if(minCurrent.right != null){
            TreeNode rightNode = minCurrent.right;
            while(rightNode != null){
                stack.push(rightNode);
                rightNode = rightNode.left;
            }
        }

       return minCurrent.val;
    }
}

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