[LeetCode]Binary Search Tree Iterator，解题报告

mplement 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.

思路

import java.util.ArrayDeque;
import java.util.Stack;


public class BSTIterator {
	private ArrayDeque<TreeNode> mArrayDeque;
	
	public BSTIterator(TreeNode root) {
		mArrayDeque = new ArrayDeque<TreeNode>();
		bTreeInorderTraverse(root);
	}
	
	private void bTreeInorderTraverse(TreeNode root) {
		TreeNode p = root;
		Stack<TreeNode> tmpStack = new Stack<TreeNode>();
		
		while (p != null || ! tmpStack.empty()) {
			if (p != null) {
				tmpStack.push(p);
				p = p.left;
			} else {
				p = tmpStack.pop();
				mArrayDeque.add(p);
				p = p.right;
			}
		}
	}
	
	public boolean hasNext() {
		return ! mArrayDeque.isEmpty();
	}
	
	public int next() {
		if (hasNext()) {
			return mArrayDeque.poll().val;
		} else {
			return -1;
		}
	}
	
	public static class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;
		TreeNode(int x) {
			val = x;
		}
	}
}

优化

1. 申请一个只有h大小的栈。
2. 在构造函数中，将该树的所有左孩子入栈。
3. 在next()函数中，弹出栈顶节点，如果该节点有右孩子，将右孩子和该右孩子的所有左孩子入栈。

import java.util.Stack;


public class BSTIterator {
	private Stack<TreeNode> mStack;
	
	public BSTIterator(TreeNode root) {
		mStack = new Stack<TreeNode>();
		while (root != null) {
			mStack.push(root);
			root = root.left;
		}
	}
	
	public boolean hasNext() {
		return ! mStack.empty();
	}
	
	public int next() {
		TreeNode p = mStack.pop();
		int res = p.val;
		if (p.right != null) {
			TreeNode node = p.right;
			while (node != null) {
				mStack.push(node);
				node = node.left;
			}
		}
		
		return res;
	}
	
	public static class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;
		TreeNode(int x) {
			val = x;
		}
	}
}