題目描述是這樣的:
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1 \ 2 / 3
return [1,2,3]
.
Note: Recursive solution is trivial, could you do it iteratively?
要求使用非遞歸方法。
可以利用棧來實現。
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public ArrayListpreorderTraversal(TreeNode root) { ArrayList list=new ArrayList (); Stack stack=new Stack(); if(root!=null){ stack.push(root); while(!stack.empty()){ list.add(stack.peek().val);//這裡只能用peek(),因為如果pop()的話會在獲得val值的同時將結點彈出棧。 root=stack.pop(); if(root.right!=null){ stack.push(root.right);//右孩子先入棧,後出棧 } if(root.left!=null){ stack.push(root.left);//左孩子後入棧,先出棧 } } } return list; } }