题目描述
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.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / \ 2 3 / 4 \ 5The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ import java.util.*; public class Solution { public ArrayList<Integer> inorderTraversal(TreeNode root) { if(root==null) { return new ArrayList<>(); } ArrayList<Integer> list=new ArrayList<>(); list.addAll(inorderTraversal(root.left)); list.add(root.val); list.addAll(inorderTraversal(root.right)); return list; } }
or:
import java.util.*; public class Solution { public ArrayList<Integer> inorderTraversal(TreeNode root) { ArrayList<Integer> res = new ArrayList<Integer>(); Stack<TreeNode> s = new Stack<TreeNode>(); while(!s.isEmpty() || root != null) { while(root != null) { s.push(root); root = root.left; } root = s.pop(); res.add(root.val); root = root.right; } return res; } }