标签:
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.
[思路]
非递归遍历BST变种
[注意]
None
[CODE]
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class BSTIterator { TreeNode node = null; Stack<TreeNode> stack = new Stack<TreeNode>(); public BSTIterator(TreeNode root) { node = root; } /** @return whether we have a next smallest number */ public boolean hasNext() { return !( (node==null) && stack.empty() ); } /** @return the next smallest number */ public int next() { TreeNode res = null; if(node==null) { res = stack.pop(); node = res.right; } else { while(node.left!=null) { stack.push(node); node = node.left; } res = node; node = node.right; } return res.val; } } /** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root);e * while (i.hasNext()) v[f()] = i.next(); */
leetcode 173: Binary Search Tree Iterator
标签:
原文地址:http://blog.csdn.net/xudli/article/details/42340627