| 标题： | Binary Tree Preorder Traversal |
| 通过率： | 36.2% |
| 难度： | 中等 |
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?
二叉树的前序遍历，直接进行递归访问即可，代码如下：
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<Integer> preorderTraversal(TreeNode root) { 12 ArrayList<Integer> result=new ArrayList<Integer>(); 13 vistTree(result,root); 14 return result; 15 } 16 public void vistTree(List<Integer> temp,TreeNode root){ 17 if(root!=null){ 18 temp.add(root.val); 19 vistTree(temp,root.left); 20 vistTree(temp,root.right); 21 } 22 } 23 }
leetcode------Binary Tree Preorder Traversal
原文：http://www.cnblogs.com/pkuYang/p/4311377.html