摘要:当你被的时候,人家问,一定要说,当然可以啦所以,不用,怎么办幸好,这是一道的题目,只要逐层遍历,就可以了。所以,试一试用堆栈的做法吧记得的特点是后入先出哦
Problem
Given a binary tree, return the preorder traversal of its nodes" values.
ExampleGiven:
1 / 2 3 / 4 5
return [1,2,4,5,3].
ChallengeCan you do it without recursion?
Note当你被challenge的时候,人家问,Can you do it without recursion? 一定要说,当然可以啦!所以,不用recursion,怎么办?幸好,这是一道preorder的题目,只要逐层遍历,就可以了。所以,试一试用堆栈的做法吧!记得Stack的特点是后入先出哦!
SolutionRecursion
public class Solution { public ArrayListpreorderTraversal(TreeNode root) { ArrayList res = new ArrayList(); if (root == null) return res; helper(res, root); return res; } public void helper(ArrayList res, TreeNode root) { res.add(root.val); if (root.left != null) helper(res, root.left); if (root.right != null) helper(res, root.right); } }
Stack
public class Solution { public ArrayListpreorderTraversal(TreeNode root) { ArrayList res = new ArrayList(); Stack stack = new Stack(); if (root == null) return res; stack.push(root); while (!stack.isEmpty()) { TreeNode cur = stack.pop(); res.add(cur.val); if (cur.right != null) stack.push(cur.right); if (cur.left != null) stack.push(cur.left); } return res; } }
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