问题描述

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},

return its level order traversal as:

代码

struct TreeNode {      int val;      TreeNode *left;      TreeNode *right;      TreeNode(int x) : val(x), left(NULL), right(NULL) {}  };vector
     
      
       > levelOrder(TreeNode* root) {       deque
       
        a;       deque
        
         b;       if (root)            a.push_back(root);       TreeNode*p = NULL;       vector
         
          
           > result; while (!a.empty()) { vector
           
            temp; if (!a.empty() ) { while (!a.empty() ) { p = a.front(); a.pop_front(); temp.push_back(p->val); if (p->left) b.push_back(p->left); if (p->right) b.push_back(p->right); } result.push_back(temp); } while (!b.empty()) { a.push_back(b.front()); b.pop_front(); } } return result; }