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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
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原文地址：http://www.cnblogs.com/andrew-chen/p/4228530.html
