leetcode Binary Tree Postorder Traversal

Binary Tree Postorder Traversal

 Total Accepted: 23892 Total Submissions: 77206My Submissions

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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第二遍写postOrder竟然有点忘记了，更惨的是preOrder竟然半个小时都没写出来。看来还是要经过思考才能够记得牢固，之前对preOrder,postOrder都是看文章然后再理解，所以有些逻辑没有好好理清楚。

下面上代码：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution { public List<Integer> postorderTraversal(TreeNode root) {        TreeNode tmp=null,node=null;        List<TreeNode> list = new ArrayList<TreeNode>();        List<Integer> res = new ArrayList<Integer>();        if(root==null)        {            return res;        }else        {            tmp = root;            // remember the add element            list.add(tmp);            // remember loop  condition            while(list.size()>0)            {                tmp = list.get(list.size()-1);

<span style="white-space:pre"></span>// the condition to add the node                if((tmp.right==null&&tmp.left==null) ||                (node!=null &&(node==tmp.left || node==tmp.right)))                {                    list.remove(list.size()-1);                    res.add(tmp.val);                    node = tmp;                }else                {                    if(tmp.right!=null)                     {                        list.add(tmp.right);                     }                    if(tmp.left!=null)                    {                        list.add(tmp.left);                    }                }            }        }        // remember the return value        return res;    }}