leetcode 173. 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.

设计实现一个带有下列属性的二叉查找树的迭代器：

元素按照递增的顺序被访问（比如中序遍历）

next()

和

hasNext()

的询问操作要求均摊时间复杂度是O(1)

用stack记录从根节点道当前节点的路径，初始化的时候要找到最左边的点，也就是中序遍历的第一个点，

为了记录这个过程，把从根节点道最左下方的节点之间的节点都压栈。

next返回的是stack栈定的元素，弹出最上面的元素后，还要看一下这个被返回的元素是否有右节点，

如果有，就把右节点及所有的左侧子节点都压入栈中。

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

public class BSTIterator {

Stack<TreeNode> stack;
public BSTIterator(TreeNode root) {
stack = new Stack<TreeNode>();
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();
int result = node.val;
if (node.right != null) {
node = node.right;
while(node != null){
stack.push(node);
node = node.left;
}
}
return result;
}
}

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