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[刷题] 102 Binary Tree Level Order Traversal

要求

对二叉树进行层序遍历

实现

返回结果为双重向量，对应树的每层元素
队列的每个元素是一个pair对，存树节点和其所在的层信息

 1 Definition for a binary tree node.
 2 struct TreeNode {
 3     int val;
 4     TreeNode *left;
 5     TreeNode *right;
 6     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 7 };
 8  
 9 class Solution {
10 public:
11     vector<vector<int>> levelOrder(TreeNode* root) {
12         vector<vector<int>> res;
13         if( root == NULL )
14             return res;
15         queue< pair<TreeNode*, int> > q;
16         q.push( make_pair( root , 0 ) );
17         while( !q.empty() ){
18             TreeNode* node = q.front().first;
19             int level = q.front().second;
20             q.pop();    
21             
22             if( level == res.size() )
23                 res.push_back( vector<int>() );
24             res[level].push_back( node->val );
25             
26             if( node->left )
27                 q.push( make_pair( node->left , level + 1 ));
28             if( node->right )
29                 q.push( make_pair( node->right , level + 1 ));    
30         } 
31         return res;
32     }
33 };

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