【leetcode】【94】Binary Tree Inorder Traversal

一、问题描述

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

二、问题分析

①采用递归的方式②采用栈来辅助，实现非递归的方式。

三、Java AC代码

public ArrayList<Integer> inorderTraversal(TreeNode root) {    ArrayList<Integer> res = new ArrayList<Integer>();    helper(root, res);    return res;}private void helper(TreeNode root, ArrayList<Integer> res){    if(root == null)        return;    helper(root.left,res);    res.add(root.val);    helper(root.right,res);}

public List<Integer> inorderTraversal(TreeNode root) {       List<Integer> list = new ArrayList<Integer>();        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();        TreeNode cur = root;        while(cur!=null||!stack.isEmpty()){        while(cur!=null){        stack.push(cur);        cur = cur.left;        }        if (!stack.isEmpty()) {list.add(stack.peek().val);cur = stack.pop().right;}        }        return list;    }