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| 标题： | Binary Tree Postorder Traversal |
| 通过率： | 31.8% |
| 难度： | 难 |
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?
二叉树的后续遍历，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> postorderTraversal(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 vistTree(temp,root.left); 19 20 vistTree(temp,root.right); 21 temp.add(root.val); 22 } 23 } 24 }
leetcode------Binary Tree Postorder Traversal
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原文地址：http://www.cnblogs.com/pkuYang/p/4311411.html
