leetcode 102二叉树的层序遍历-LMLPHP

5月20更新:

使用借助队列实现bfs,定义len记录队列的尺寸直接进行遍历层序

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/**
采用一个队列,使用一个变量记录上一层的元素个数。
**/ class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res;
if(root==NULL) return res;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()){
vector<int> level;
int len=q.size();
for(int i=;i<len;i++){
TreeNode* p=q.front();
q.pop();
level.push_back(p->val);
if(p->left!=NULL) q.push(p->left);
if(p->right!=NULL) q.push(p->right);
}
res.push_back(level);
}
return res;
}
};


更新之前:使用广度优先搜索和获得队列大小:

/**
* 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) {
if(root==NULL) return {};
queue<TreeNode*> q;
TreeNode* front;
q.push(root);
vector<vector<int>> res; while(!q.empty()){
vector<int> onelevel;
for(int i=q.size();i>;i--){
front=q.front();
q.pop();
if(front->left)
q.push(front->left);
if(front->right)
q.push(front->right);
onelevel.push_back(front->val);
}
res.push_back(onelevel);
}
return res;
}
};

使用两个队列:

/**
* 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>> res;
queue<TreeNode*> q1;
queue<TreeNode*> q2;
if(root==NULL) return {};
q1.push(root);
while(!q1.empty()){
vector<int> level;
while(!q1.empty()){
TreeNode* cur=q1.front();
q1.pop();
level.push_back(cur->val);
if(cur->left)
q2.push(cur->left);
if(cur->right)
q2.push(cur->right);
}
while(!q2.empty()){
q1.push(q2.front());
q2.pop();
}
res.push_back(level);
}
return res;
}
};
05-27 23:54