[leetcode] 102. Binary Tree Leve

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]
]

解题思路：
本题让我们求解数的宽度优先遍历，具体思路有使用队列将每一层的节点保存下来，然后按照FIFO的选择逐个遍历，然后将每个节点的子节点存入队列，以此类推。

具体代码如下:

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ret;
        queue<TreeNode*> level;
        if(!root) return ret;
        level.push(root);
        
        while(!level.empty())
        {
            vector<int> curLevel;
            int size = level.size();
            for(int i = 0; i < size; ++i)
            {
                TreeNode* curr = level.front();
                curLevel.push_back(curr->val);
                level.pop();
                if(curr->left) level.push(curr->left);
                if(curr->right) level.push(curr->right);
            }
            ret.push_back(curLevel);
        }
        return ret;
    }
};