LeetCode 102.Binary Tree Level Order Traversal (二叉树的层次遍历)

题目描述：

给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。

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]
]
      

AC C++ Solution:

递归：

/**
 * 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) {
        buildvector(root,0);
        return res;
    }
    
    void buildvector(TreeNode *root, int depth) 
    {
        if(root == NULL)    return;
        if(res.size() == depth)
            res.push_back(vector<int>());
        
        res[depth].push_back(root->val);
        buildvector(root->left, depth+1);
        buildvector(root->right,depth+1);
    }

private:
    vector< vector<int> > res;
};
           

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;
        
        if(!root)   return res;
        
        queue<TreeNode*> q;
        q.push(root);
        q.push(NULL);
        vector<int> cur_vec;
        
        while(!q.empty()) {
            TreeNode *t = q.front();
            q.pop();
            
            if(t == NULL) {
                res.push_back(cur_vec);
                cur_vec.resize(0);
                if(q.size() > 0)
                    q.push(NULL);       //每层之间用一个NULL作为分隔符
            }
            else {                      //把每个节点的值存入cur_vec,并把下一层的节点加入队列
                cur_vec.push_back(t->val);  
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
        }
        return res;
    }
};