N-ary Tree Level Order Traversal
2011 年 7 月 19 日
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:
Input: root = [1,null,3,2,4,null,5,6] Output: [[1],[3,2,4],[5,6]]
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],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
Constraints:
1000 [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。
class Node {
public:
int val;
vector children;
Node() {}
Node(int _val, vector _children) {
val = _val;
children = _children;
}
};
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：
- DFS
vector<vector> res;
vector<vector> levelOrder(Node* root) {
dfsWithHeight(root, 0);
return res;
}
void dfsWithHeight(Node *root, int h) {
if (root == nullptr) return;
if (h == res.size()) res.push_back(vector());
res[h].push_back(root->val);
for(int i = 0;i children.size(); i++) {
dfsWithHeight(root->children[i], h + 1);
}
}
- BFS
vector<vector> levelOrder2(Node* root) {
vector<vector> res;
if (root == nullptr) {
return res;
}
vector vec;
queue q;
q.push(root);
q.push(nullptr);
while(q.size() != 1) {
vec.clear();
root = q.front();
while(root) {
q.pop();
vec.push_back(root->val);
// cout <val << endl;
for(int i = 0;i children.size(); i++) {
q.push(root->children[i]);
}
root = q.front();
}
q.pop();
q.push(nullptr);
res.push_back(vec);
}
return res;
}
vector<vector> levelOrder1(Node* root) {
vector<vector> res;
if (root == nullptr) {
return res;
}
vector vec;
vector nodes;
vector nextLevelNodes;
nodes.push_back(root);
while(!nodes.empty()) {
vec.clear();
nextLevelNodes.clear();
for(int i = 0;i val);
for(int j = 0;j children.size(); j++) {
nextLevelNodes.push_back(nodes[i]->children[j]);
}
}
swap(nodes, nextLevelNodes);
res.push_back(vec);
}
return res;
}
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