标签:tor its -- iter bsp 技术 size code sse
题目:
Given an n-ary tree, return the preorder traversal of its nodes‘ values.
For example, given a 3-ary
tree:
Return its preorder traversal as: [1,3,5,6,2,4]
.
Note:
Recursive solution is trivial, could you do it iteratively?
/* // Definition for a Node. class Node { public: int val; vector<Node*> children; Node() {} Node(int _val, vector<Node*> _children) { val = _val; children = _children; } }; */ class Solution { public: vector<int> preorder(Node* root) { vector<int> order = {}; if (root) { traversal(root, order); } return order; } void traversal(Node* root, vector<int> &order) { order.push_back(root->val); int num = root->children.size(); for (int i = 0; i < num; i++) { traversal(root->children.at(i), order); } } };
2. Iterative solution:
/* // Definition for a Node. class Node { public: int val; vector<Node*> children; Node() {} Node(int _val, vector<Node*> _children) { val = _val; children = _children; } }; */ class Solution { public: vector<int> preorder(Node* root) { vector<int> order = {}; stack<Node*> nodeStack; if (root) { nodeStack.push(root); } while (!nodeStack.empty()) { Node* node = nodeStack.top(); nodeStack.pop(); order.push_back(node->val); int num = node->children.size(); for (int i = num - 1; i >= 0; i--) { nodeStack.push(node->children.at(i)); } } return order; } };
【leetcode】589. N-ary Tree Preorder Traversal
标签:tor its -- iter bsp 技术 size code sse
原文地址:https://www.cnblogs.com/AndrewGhost/p/10348109.html