第（6）题 binary-tree-postorder-traversa 知识点：树 题述：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]. 题意是用后根法遍历二叉树 思路：后根法要求先左子结点，再右子结点，最后再根结点的顺序遍历二叉树。使用递归会很方便。 代码如下：
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */import java.util.ArrayList;public class Solution { public ArrayList<Integer> postorderTraversal(TreeNode root) { ArrayList<Integer> list = new ArrayList<Integer>(); postorder(root, list); return list; } public void postorder(TreeNode root, ArrayList<Integer> list){ if(root == null){ return; } if(root.left != null){ postorder(root.left, list); } if(root.right != null){ postorder(root.right, list); } list.add(root.val); }}新闻热点
疑难解答
