102. 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]
]

BFS解法；

/** * 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>> ans;        if (root == NULL)        {            return ans;        }        queue<TreeNode*> q;        q.push(root);        while (!q.empty())        {            vector<int> tmp;            int size = q.size();            for (int i = 0; i < size; i++) //这个地方比较巧妙，记录q在每层的大小            {                TreeNode* node = q.front();                q.pop();                tmp.push_back(node->val);                if (node->left != NULL) q.push(node->left);                if (node->right != NULL) q.push(node->right);            }            ans.push_back(tmp);        }        return ans;    }};

DFS的解法，
http://www.jiuzhang.com/solutions/binary-tree-level-order-traversal/

每次都要从根节点遍历到第K层，时间复杂度很高。