【LeetCode】Binary Tree Level Order Traversal

地址：http://oj.leetcode.com/problems/binary-tree-level-order-traversal/#

Binary Tree Level Order Traversal

  Total Accepted: 7502  Total Submissions: 24956 My Submissions

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},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as  "{1,2,3,#,#,4,#,#,5}".

Discuss

#我的答案

/**
 * Definition for binary tree
 * 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) {
        int cur=0, cnt=0; 
        TreeNode *pnode;
        vector<vector<int>> vv;
        vector<int> vi;          // 保存每层的输出节点值
        vector<int> vn;          // 记录每层含有的节点数
        queue<TreeNode*> q;
        
        vn.push_back(1);         // level 0 有一个节点
        q.push(root);
        
        if (root != 0) while(!q.empty()){       // 广度优先搜索
            pnode = q.front();
            q.pop();
            vi.push_back(pnode->val);

            if (++cnt == 1)        // 输出的是当前层第一个节点
                vn.push_back(0);   // 为存储下一层节点开辟空间
            
            if (pnode->left != 0){
                q.push(pnode->left);
                vn[cur+1]++;
            }
            if (pnode->right != 0){
                q.push(pnode->right);
                vn[cur+1]++;
            }
            
            if (cnt == vn[cur]){  // 当前level所有节点输出完毕，保存结果并更新相关计数
                cur++;
                cnt = 0;
                vv.push_back(vi);
                vi.clear();
            }
        }
        return vv;
    }
};