给你二叉树的根节点 root ,返回它节点值的前序遍历。
示例 1:
输入:root = [1,null,2,3]
输出:[1,2,3]
示例 2:
输入:root = []
输出:[]
示例3:
输入:root = [1]
输出:[1]
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
// 使用栈
//尽量使用Deque来创建stack,不要使用默认的satck,那是一个过时的集合
Deque<TreeNode> stack = new LinkedList<>();
//创建list用于接收返回值
List<Integer> list = new ArrayList<>();
if (root == null) return list;
stack.push(root);
while(!stack.isEmpty())
{
TreeNode temp = stack.pop();
list.add(temp.val);
if (temp.right != null)
{
//先添加右子树节点,因为这样出栈的时候顺序才是前序顺序
stack.push(temp.right);
}
if (temp.left != null)
{
stack.push(temp.left);
}
}
return list;
}
}