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   7return its level order traversal as:[  [3],  [9,20],  [15,7]]

bfs层序遍历整颗树。

/** * 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
    
     
      > levelOrder(TreeNode* root) {        queue
      
        q;        vector
       
        
          > ans;        if (root == nullptr) return ans;        q.push(root);        while (!q.empty()) {            vector
         
           v; int n = q.size(); while (n-- > 0) { TreeNode* u = q.front(); q.pop(); //cout << u->val << endl; v.push_back(u->val); if (u->left != nullptr) q.push(u->left); if (u->right != nullptr) q.push(u->right); } ans.push_back(v); } return ans; }};