leetcode 102二叉树的层序遍历

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;
    }
};

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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;
    }
};