LeetCode--Binary Tree Level Order Traversal

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]]

方法一：递归。层次遍历，先遍历左节点，再遍历右节点，每一层用一个动态数组存储数据。

/** * 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>>result;        traverse(root,1,result);        return result;    }    void traverse(TreeNode* root,int level,vector<vector<int>>&result){        if(!root) return;        if(level>result.size()){            result.push_back(vector<int>());        }        result[level-1].push_back(root->val);        traverse(root->left,level+1,result);        traverse(root->right,level+1,result);    }};

方法二：非递归。用一个current队列，遍历这一层数据，存到动态数组中，再用一个next队列，存储下一层数据，然后交换到current队列，循环进行。

/** * 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>>result;        queue<TreeNode*>current,next;        if(!root) return result;        else current.push(root);        while(!current.empty()){            vector<int>level;            while(!current.empty()){                TreeNode *node=current.front();                current.pop();                    level.push_back(node->val);                if(node->left) next.push(node->left);                if(node->right) next.push(node->right);            }            result.push_back(level);            swap(current,next);        }        return result;    }};