Code & Func

第19天。

今天的题目是 N-ary Tree Level Order Traversal :

Given an n-ary tree, return the level order 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:

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

Example 2:

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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]
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Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

Constraints:

The height of the n-ary tree is less than or equal to 1000
The total number of nodes is between [0, 10^4]

一道水题，简单的BFS或DFS即可，除了是一个多叉树外，和另外一道题基本是一样的。

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

所以，我们既可以用队列去做层次遍历(BFS)，也可以用递归来实现DFS，然后按当前节点所在的高度插入到对于的数组即可：

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vector<vector<int>> res;
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vector<vector<int>> levelOrder(Node* root) {
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    dfsWithHeight(root, 0);
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    return res;
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}
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void dfsWithHeight(Node *root, int h) {
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    if (root == nullptr) return;
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    if (h == res.size()) res.push_back(vector<int>());
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    res[h].push_back(root->val);
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    for(int i = 0;i < root->children.size(); i++) {
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        dfsWithHeight(root->children[i], h + 1);
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    }
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}

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vector<vector<int>> levelOrder2(Node* root) {
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    vector<vector<int>> res;
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    if (root == nullptr) {
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        return res;
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    }
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    vector<int> vec;
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    queue<Node *> q;
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    q.push(root);
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    q.push(nullptr);
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    while(q.size() != 1) {
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        vec.clear();
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        root = q.front();
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        while(root) {
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            q.pop();
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            vec.push_back(root->val);
37 collapsed lines
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            // cout << root->val << endl;
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            for(int i = 0;i < root->children.size(); i++) {
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                q.push(root->children[i]);
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            }
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            root = q.front();
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        }
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        q.pop();
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        q.push(nullptr);
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        res.push_back(vec);
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    }
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    return res;
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}
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vector<vector<int>> levelOrder1(Node* root) {
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    vector<vector<int>> res;
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    if (root == nullptr) {
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        return res;
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    }
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    vector<int> vec;
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    vector<Node *> nodes;
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    vector<Node *> nextLevelNodes;
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    nodes.push_back(root);
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    while(!nodes.empty()) {
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        vec.clear();
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        nextLevelNodes.clear();
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        for(int i = 0;i < nodes.size(); i++) {
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            vec.push_back(nodes[i]->val);
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            for(int j = 0;j < nodes[i]->children.size(); j++) {
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                nextLevelNodes.push_back(nodes[i]->children[j]);
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            }
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        }
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        swap(nodes, nextLevelNodes);
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        res.push_back(vec);
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    }
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    return res;
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}