标签：front result blog 题目 ott root pop 数据 bottom
/*
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) {
queue<TreeNode*> q;
q.push(root);
vector<int> r;
while(!q.empty()) {
root=q.front();
q.pop();
if(!root) continue;
r.push_back(root->val);
q.push(root->left);
q.push(root->right);
}
return r;
}
};
/*C++ 循环 实现要打印成一行，即用动态数组存储即可这道题考的是广度优先遍历算法，这个算法是用队列这种数据结构实现的。STL中的deque是“两端都可以进出”的队列，queue只能是“后端进前端出”的队列。这里选用deque和queue都行，但我选择使用deque尝试一下“高大上”的队列。*//*
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) {
deque<TreeNode*> dequeNode;
vector<int> result;
if(root==NULL) return result;
dequeNode.push_back(root);
while(dequeNode.size()) {
TreeNode * pNode=dequeNode.front();
result.push_back(pNode->val);
dequeNode.pop_front();
if(pNode->left!=NULL)
dequeNode.push_back(pNode->left);
if(pNode->right!=NULL)
dequeNode.push_back(pNode->right);
}
return result;
}
};
标签：front result blog 题目 ott root pop 数据 bottom
原文地址：http://www.cnblogs.com/dd2hm/p/7299505.html
