[LeetCode]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,#,#,15,7},

    3
   / 
  9  20
    /  
   15   7

return its level order traversal as:

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

这道题目是二叉树的层次遍历。

思路：借助于队列结构(queue)，保存尚未访问儿子节点的节点。

这道题一大注意点在于，每层结束后插入NULL作为分隔符。

1. 初始化：将根节点和第一层终止符NULL压入队列

2. 循环

用level记录该层的节点val

- 取出队首元素cur

- 若cur非空, 记录cur->val

  若cur为NULL，则说明该层节点已全部访问过，将level插入ret. 

  若此时queue非空，访问下一层节点，否则，停止循环

代码

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<vector<int>> ret;
        if(root==NULL)
            return ret;
        
        queue<TreeNode *> q;
        q.push(root);
        q.push(NULL); // IMPORTANT! end of level
        
        vector<int> level;
        
        while(!q.empty())
        {
            TreeNode *cur = q.front();
            q.pop();
            
            if(cur)
            {
                level.push_back(cur->val);
                if(cur->left)
                    q.push(cur->left);
                if(cur->right)
                    q.push(cur->right);
            }
            else
            {
                ret.push_back(level);
                level.clear();
                if(q.empty())
                    break;
                else
                    q.push(NULL);
            }
        }
        return ret;
    }
};