题目描述
从上往下打印出二叉树的每个节点,同层节点从左至右打印。
因为对STL的队列不熟悉,这里用vector来实现队列的功能:先进先出
下面有用队列实现的,更加简洁
/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};*/
class Solution {
public:
vector<int> PrintFromTopToBottom(TreeNode* root) {
//存储要打印的每个节点的内容
vector<int> printVec;
//存储要打印的每个节点的指针
vector<TreeNode*> nodeVec;
if(root == nullptr)
return printVec;
//将根节点加入
nodeVec.push_back(root);
//若当前节点有左子节点或者右子节点,则将其加入(先左后右),同时将当前节点的内容加入printVec
//然后指向nodeVec中的下一个节点
for(int i = 0; i < nodeVec.size(); i++){
if(nodeVec[i]->left != nullptr)
nodeVec.push_back(nodeVec[i]->left);
if(nodeVec[i]->right != nullptr)
nodeVec.push_back(nodeVec[i]->right);
printVec.push_back(nodeVec[i]->val);
}
return printVec;
}
};
用队列queue实现,更简洁:
/*
链接:https://www.nowcoder.com/questionTerminal/7fe2212963db4790b57431d9ed259701
来源:牛客网
*/
class Solution {
public:
vector<int> PrintFromTopToBottom(TreeNode* root) {
vector<int> res;
if(root==NULL)
return res;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()){
res.push_back(q.front()->val);
if(q.front()->left!=NULL)
q.push(q.front()->left);
if(q.front()->right!=NULL)
q.push(q.front()->right);
q.pop();
}
return res;
}
};