Leetcode-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 do it iteratively?

Solution:

1 /**2 * Definition for binary tree3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  Public classSolution { One      PublicList<integer>postordertraversal (TreeNode root) { Alist<integer> res =NewArraylist<integer>(); -         if(root==NULL)returnRes; -  theStack<treenode> Nstack =NewStack<treenode>(); -Stack<integer> record =NewStack<integer>(); - Nstack.push (root); -Record.push (0); +  -          while(Nstack.size ()!=0){ +TreeNode cur =Nstack.peek (); A             if(cur.left==NULL&& cur.right==NULL){ at Res.add (cur.val); - Nstack.pop (); - Record.pop (); -                 Continue; -             } -  in             intval =Record.peek (); -             if(val==0){ to Record.pop (); +Record.push (val+1); -                 if(cur.left!=NULL){ the Nstack.push (cur.left); *Record.push (0); $                 }Panax Notoginseng                 Continue; -             } the  +             if(val==1){ A Record.pop (); theRecord.push (val+1); +                 if(cur.right!=NULL){ - Nstack.push (cur.right); $Record.push (0); $                 } -                 Continue; -             } the  -             if(val==2){Wuyi Res.add (cur.val); the Nstack.pop (); - Record.pop (); Wu                 Continue; -             } About         } $  -         returnRes;  -          -     } A}

Leetcode-binary Tree postorder Traversal