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.
Example:
BSTIterator iterator = new BSTIterator(root); iterator.next();// return 3 iterator.next();// return 7 iterator.hasNext(); // return true iterator.next();// return 9 iterator.hasNext(); // return true iterator.next();// return 15 iterator.hasNext(); // return true iterator.next();// return 20 iterator.hasNext(); // return false
Note:
next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.
難度:medium
題目:實現二叉搜尋樹的遍歷。迭代器用根結點初始化。呼叫next()將會返回下一個最小的結點。
Runtime: 76 ms, faster than 37.23% of Java online submissions for Binary Search Tree Iterator.
Memory Usage: 52.1 MB, less than 100.00% of Java online submissions for Binary Search Tree Iterator.
/** * Definition for a binary tree node. * public class TreeNode { *int val; *TreeNode left; *TreeNode right; *TreeNode(int x) { val = x; } * } */ class BSTIterator { private Stack<TreeNode> stack; public BSTIterator(TreeNode root) { stack = new Stack<>(); while (root != null) { stack.push(root); root = root.left; } } /** @return the next smallest number */ public int next() { TreeNode node = stack.pop(); TreeNode ptr = node.right; while (ptr != null) { stack.push(ptr); ptr = ptr.left; } return node.val; } /** @return whether we have a next smallest number */ public boolean hasNext() { return !stack.isEmpty(); } } /** * Your BSTIterator object will be instantiated and called as such: * BSTIterator obj = new BSTIterator(root); * int param_1 = obj.next(); * boolean param_2 = obj.hasNext(); */