Binary Tree Postorder Traversal
Given a binary tree, return the postorder traversal of its nodes‘ values.
For example:
Given binary tree{1,#,2,3},1 2 / 3
return
[3,2,1].Note: Recursive solution is trivial, could you do it iteratively?
算法思路：
最简单的后序遍历。不解释了。
代码如下：
1 public class Solution { 2 List<Integer> res = new ArrayList<Integer>(); 3 public List<Integer> postorderTraversal(TreeNode root) { 4 if(root == null) return res; 5 if(root.left != null) postorderTraversal(root.left); 6 if(root.right != null) postorderTraversal(root.right); 7 res.add(root.val); 8 return res; 9 } 10 }
[leetcode]Binary Tree Postorder Traversal,布布扣,bubuko.com
[leetcode]Binary Tree Postorder Traversal
原文：http://www.cnblogs.com/huntfor/p/3900239.html