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?
Solution:
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 List<Integer> res = new ArrayList<Integer>(); 13 if (root==null) return res; 14 15 Stack<TreeNode> nStack = new Stack<TreeNode>(); 16 Stack<Integer> record = new Stack<Integer>(); 17 nStack.push(root); 18 record.push(0); 19 20 while (nStack.size()!=0){ 21 TreeNode cur = nStack.peek(); 22 if (cur.left==null && cur.right==null){ 23 res.add(cur.val); 24 nStack.pop(); 25 record.pop(); 26 continue; 27 } 28 29 int val = record.peek(); 30 if (val==0){ 31 record.pop(); 32 record.push(val+1); 33 if (cur.left!=null){ 34 nStack.push(cur.left); 35 record.push(0); 36 } 37 continue; 38 } 39 40 if (val==1){ 41 record.pop(); 42 record.push(val+1); 43 if (cur.right!=null){ 44 nStack.push(cur.right); 45 record.push(0); 46 } 47 continue; 48 } 49 50 if (val==2){ 51 res.add(cur.val); 52 nStack.pop(); 53 record.pop(); 54 continue; 55 } 56 } 57 58 return res; 59 60 } 61 }