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?
求前序遍历,要求不用递归。
使用双向队列。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> preorderTraversal(TreeNode root) { Deque<TreeNode> queue = new ArrayDeque<TreeNode>(); List list = new ArrayList<Integer>(); if( root == null ) return list; queue.add(root); while( !queue.isEmpty() ){ TreeNode node = queue.poll(); list.add(node.val); if( node.right != null ) queue.addFirst(node.right); if( node.left != null ) queue.addFirst(node.left); } return list; } }