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].
递归遍历比较简单，代码如下：
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { List<Integer> list = new ArrayList<Integer>(); public void POT(TreeNode root) { if(root!=null) { if(root.left!=null) POT(root.left); if(root.right!=null) POT(root.right); list.add(root.val); } } public List<Integer> postorderTraversal(TreeNode root) { POT(root); return list; } }
Binary Tree Postorder Traversal
原文：http://www.cnblogs.com/mrpod2g/p/4364484.html