LeetCode 102. 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,null,null,15,7] ,

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

解析：二叉树层次遍历，输出格式按照从左到右，分层换行输出。对于二叉树层次遍历采用队列的形式，然后每层遍历的时候记录当前层节点数量。分析汇总参见之前的博客

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if(root == NULL)
            return result;
        queue<TreeNode*> que;
        que.push(root);
        while(!que.empty())
        {
            int len = que.size();
            vector<int> level_result;
            for(int i=0; i < len; i++)
            {
                TreeNode* node = que.front();
                que.pop();
                level_result.push_back(node->val);
                if(node->left)
                    que.push(node->left);
                if(node->right)
                    que.push(node->right);
            }
            result.push_back(level_result);
        }
        return result;
    }
};