LeetCode Binary Search Tree Iterator

原题链接在这里：https://leetcode.com/problems/binary-search-tree-iterator/

题目：

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：

用一个stack把所有最左边的node压进 stack中。

判断hasNext()时就是看stack 是否为空.

next()返回stack顶部top元素，若是top有右子树，就把右子树的所有最左node压入stack中.

constructor time complexity: O(h).

hashNext time complexity: O(1).

next time complexity: O(h).

Space: O(h), stack 最大为树的高度。

AC Java:

 /**

  * Definition for binary tree

  * public class TreeNode {

  *     int val;

  *     TreeNode left;

  *     TreeNode right;

  *     TreeNode(int x) { val = x; }

  * }

  */

 public class BSTIterator {

     Stack<TreeNode> stk;

     public BSTIterator(TreeNode root) {

         stk = new Stack<TreeNode>();

         while(root != null){

             stk.push(root);

             root = root.left;

         }

     }

     /** @return whether we have a next smallest number */

     public boolean hasNext() {

         return !stk.isEmpty();

     }

     /** @return the next smallest number */

     public int next() {

         TreeNode top = stk.pop();

         TreeNode rightLeft = top.right;

         while(rightLeft != null){

             stk.push(rightLeft);

             rightLeft = rightLeft.left;

         }

         return top.val;

     }

 }

 /**

  * Your BSTIterator will be called like this:

  * BSTIterator i = new BSTIterator(root);

  * while (i.hasNext()) v[f()] = i.next();

  */

类似Binary Tree Inorder Traversal, Inorder Successor in BST.

巴特西

LeetCode Binary Search Tree Iterator

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