Given an n-ary tree, return the preorder traversal of its nodes' values.
Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
Follow up:
Recursive solution is trivial, could you do it iteratively?
Example 1:
Input: root = [1,null,3,2,4,null,5,6] Output: [1,3,5,6,2,4]
Example 2:
Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14] Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]
Constraints:
1000
[0, 10^4]
N叉树的前序遍历。刚听完覃超老师讲的二叉树。
所以成热打铁练习下easy级别的。
就是采用递归的方式:先根,再从左到右遍历子节点。
public List preorder(Node root) {
if (root != null){
list.add(root.val);
for(int i=0;i< root.children.size();i++ ) {
preorder(root.children.get(i));
}
}
return list;
}