LeetCode Binary Tree Level Order Traversal

class Solution {
private:
    vector<vector<int> > nodes;
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        nodes.clear();
        dfs(root, 0);
        return nodes;
    }
    
    void dfs(TreeNode* root, int level) {
        if (root == NULL) return;
        if (level >= nodes.size()) {
            nodes.push_back(vector<int>());
        }
        nodes[level].push_back(root->val);
        dfs(root->left, level + 1);
        dfs(root->right, level + 1);
    }
};

水一发

第二轮：

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / 
  9  20
    /  
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

/**
 * Definition for binary tree
 * 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;
        //dfs(res, root, 0);
        bfs(res, root);
        return res;
    }
    void dfs(vector<vector<int> >& res, TreeNode* root, int level) {
        if (root == NULL) {
            return;
        }
        if (level == res.size()) {
            res.push_back(vector<int>());
        }
        res[level].push_back(root->val);
        dfs(res, root->left, level + 1);
        dfs(res, root->right, level + 1);
    }
    
    void bfs(vector<vector<int> >& res, TreeNode* root) {
        if (root == NULL) {
            return;
        }

        
        queue<TreeNode*> que;
        que.push(root);
        
        while (!que.empty()) {
            int n = que.size();
            vector<int> current;
            for (int i=0; i<n; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (node == NULL) {
                    continue;
                }
                current.push_back(node->val);
                if (node->left != NULL) {
                    que.push(node->left);
                }
                if (node->right != NULL) {
                    que.push(node->right);
                }
            }
            res.push_back(current);
        }
    } 
};