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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?
1 /** 2 * Definition for a binary tree node. 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 List<Integer> result = new ArrayList<Integer>(); 12 public List<Integer> preorderTraversal(TreeNode root) { 13 preorderTraversal2(root); 14 return result; 15 } 16 17 public void preorderTraversal2(TreeNode root){ 18 if (root == null) 19 return; 20 result.add(root.val); 21 preorderTraversal(root.left); 22 preorderTraversal(root.right); 23 } 24 }
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原文地址:http://www.cnblogs.com/timoBlog/p/4782853.html