LeetcCode102 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). （Easy）

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]
]

分析：

层序遍历，就是BFS的思路，利用队列把每一行的节点存进去，然后一行一行读出。

注意每次循环用读取size的方式确定这一行到哪里结束。

代码：

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