作者:whisper
链接:https://www.proprogrammar.com/article/961
声明:请尊重原作者的劳动,如需转载请注明出处
从上到下按层打印二叉树,同一层的节点按从左到右的顺序打印,每一层打印到一行。
例如:
给定二叉树: [3,9,20,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
返回其层次遍历结果:
[ [3], [9,20], [15,7] ]
提示:
节点总数 <= 1000
难度:简单;标签:树,广度优先搜索;编程语言:C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
vector<TreeNode*> cur;
vector<TreeNode*> next;
TreeNode* t;
vector<int> temp;
public:
// c++重写
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res;
if(!root)return res;
cur.push_back(root);
while(!cur.empty()){
while(!cur.empty()){
t = cur.front();
cur.erase(cur.begin());
temp.push_back(t->val);
if(t->left)next.push_back(t->left);
if(t->right)next.push_back(t->right);
}
vector<int> add;
temp.swap(add);
res.push_back(add);
cur.swap(next);
}
return res;
}
};
分两层,当前层和下一层,每一轮遍历保存当前层,同时获取下一层结点,也可以用队列完成
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> vecReturn;
// vecFloor.push_back(root->val);
if(root)
{
queue<TreeNode*> queFloor;
queFloor.push(root);
while(!queFloor.empty())
{
int nSize = queFloor.size();
vector<int> vecFloorChild;
for(int i=0;i<nSize;++i)
{
TreeNode* pCur = queFloor.front();
vecFloorChild.push_back(pCur->val);
queFloor.pop();
if(pCur->left)
{
queFloor.push(pCur->left);
}
if(pCur->right)
{
queFloor.push(pCur->right);
}
}
vecReturn.push_back(vecFloorChild);
}
}
return vecReturn;
}
};
上面是队列实现的代码,每一轮要保存当前的size
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