LeetCode: Binary Tree Postorder Traversal 解题报告

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?

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LeetCode: Binary Tree Postorder Traversal 解题报告

SOLUTION 1:

递归解法

 1 public List<Integer> postorderTraversal1(TreeNode root) {
 2         List<Integer> ret = new ArrayList<Integer>();
 3         dfs(root, ret);
 4         return ret;
 5     }
 6     
 7     // Solution 1: rec
 8     public void dfs(TreeNode root, List<Integer> ret) {
 9         if (root == null) {
10             return;
11         }
12         
13         dfs(root.left, ret);
14         dfs(root.right, ret);
15         ret.add(root.val);
16     }

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