LeetCode – Binary Tree Level Order Traversal Solution

题目：

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

解题：

       两种方式处理， 广度优先搜索借助队列，一层一层处理，并放入二维数组； 深度优先搜索借助栈， 一个一个处理， 将每一个元素方式所属层对应的一维数组中。 

实现：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        
   
        vector<vector<int>> result;
        
        if (!root) {
            return result;
        }
               
        queue<TreeNode*> q;
        q.push(root);
        
        while (!q.empty()){
            
            vector<int> current;       
            int level_size = q.size();
            
            for (int i=0; i<level_size; i++){
                 TreeNode* node = q.front();
                 q.pop();
                 current.push_back(node->val);
                
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
            
            result.push_back(current);
            
        }
        
        return result;
    }
};

实现：

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if (!root){
            return result;
        }
        result = _dfs(root, 0, result);
        return result;
    }
    
        
    vector<vector<int>> _dfs(TreeNode *node, int level, vector<vector<int>> &res){
        if (!node){
           return res;
        }
        if (res.size() < level+1){
            res.push_back({});
        }
        res[level].push_back(node->val);
        _dfs(node->left, level+1, res);
        _dfs(node->right, level+1, res);
        
        return res;
    }
};