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LeetCode 解题报告(351) -- 589. N-ary Tree Preorder Traversal

Problem

Given the root of 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)

Example 1:

Example1

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Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]

Example 2:

Example2

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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:

  • The number of nodes in the tree is in the range [0, 104].
  • 0 <= Node.val <= 104
  • The height of the n-ary tree is less than or equal to 1000.

Follow up: Recursive solution is trivial, could you do it iteratively?


Analysis

  N 叉树的前序遍历,树的遍历无非就是递归和循环。对于二叉树来说,递归的解法更加简单清晰,但是对于 N 叉树来说,使用循环可能比较容易。其实思路与二叉树的一样的,还是借助栈来做。

   每处理一个节点时,把它的子节点都压入栈中,但是这里注意顺序,因为遍历的顺序是从左到右,所以压入的顺序是从右到左。按照这个逻辑处理直至栈中没有节点即可。


Solution

   无


Code

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/*
// Definition for a Node.
class Node {
public:
int val;
vector<Node*> children;

Node () {}

Node (int _val) {
val = _val;
}

Node (int _val, vector<Node*> _children) {
val = _val;
children = _children;
}
};
*/

class Solution {
public:
vector<int> preorder(Node* root) {
vector<int> result;
stack<Node*> s;
if (!root) {
return result;
}

s.push (root);
while (!s.empty ()) {
Node* node = s.top ();
s.pop ();
result.push_back (node->val);
int size = node->children.size ();
for (int i = size - 1; i >= 0; i--) {
s.push (node->children [i]);
}
}
return result;
}
};

Summary

   这道题目本质上考察的是二叉树的前序遍历非递归实现,原理是使用栈来解决。N 叉树和二叉树处理起来没有什么不同。这道题目的分享到这里,感谢你的支持!

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