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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?
/** * 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) { List<Integer> list=new ArrayList<Integer>(); post(root,list); return list; } public void post(TreeNode node,List<Integer> list){ if(node==null) return; post(node.left,list); post(node.right,list); list.add(node.val); } }思路:二叉树的递归后序遍历,非递归的也得会。
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