首页 > 代码库 > [Leetcode] Binary Tree Preorder Traversal
[Leetcode] Binary Tree Preorder Traversal
Given a binary tree, return the preorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3}
,
1 2 / 3
return [1,2,3]
.
Note: Recursive solution is trivial, could you do it iteratively?
[Thoughts]
常见面试题,主要考两个东西:
1.什么是先序遍历
2.Java中Stack的使用
[Code]
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { public List<Integer> preorderTraversal(TreeNode root) { List<Integer> result = new ArrayList<Integer>(); if(root == null) return result; Stack<TreeNode> stack = new Stack<TreeNode>(); stack.push(root); while(!stack.isEmpty()){ TreeNode temp = stack.pop(); result.add(temp.val); if(temp.right != null){ stack.push(temp.right); } if(temp.left != null){ stack.push(temp.left); } } return result; }}
[Leetcode] Binary Tree Preorder Traversal
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。