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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.
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class BSTIterator { Stack<TreeNode> stack=new Stack<TreeNode>(); //中序遍历+stack //非递归中序遍历二叉树的话就是借助栈,让下次输出的目标节点始终存放在栈顶位置。每次输出一个节点, //就将此节点的后续节点放入栈中(沿着右子树的左子树一直向左就是下一个要输出的节点)。 public BSTIterator(TreeNode root) { pushStack(root); } /** @return whether we have a next smallest number */ public boolean hasNext() { return !stack.empty(); } /** @return the next smallest number */ public int next() { TreeNode node=stack.pop(); pushStack(node.right);///注意保存右节点 return node.val; } private void pushStack(TreeNode node){ while(node!=null){ stack.push(node); node=node.left; } } } /** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */
[leedcode 173] Binary Search Tree Iterator
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原文地址:http://www.cnblogs.com/qiaomu/p/4696398.html