[LeetCode刷题日记(Day23)]:Binary Tree Level Order Traversal-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_36348299/article/details/89020936

本文介绍了一种经典的二叉树层次遍历算法，并通过C++和Python代码实现了该算法。利用广度优先搜索（BFS）策略，通过队列记录每一层节点的值，最终得到从上到下、从左到右的层次遍历结果。

Problem 102. Binary Tree Level Order Traversal

题目描述
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
仔细观察可以发现，本题就是 BFS（广度优先搜索）最好的例子了，可用队列 queue 来实现此算法（在 python 下可直接用 list 充当 queue）。
代码实现
以下分别是 C++ 和 Python 的代码

class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode* root) {
        vector<vector<int> > res;
        if(!root)
            return res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            vector<int> level;
            int size = q.size();
            while(size--){
                TreeNode* cur = q.front();
                q.pop();
                level.push_back(cur->val);
                if(cur->left)
                    q.push(cur->left);
                if(cur->right)
                    q.push(cur->right);
            }
            res.push_back(level);
        }
        return res;
    }
};

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        res = []
        if root == None:
            return res
        queue = [root]
        while(len(queue)):
            level = []
            for i in range(len(queue)):
                cur = queue.pop(0)
                level.append(cur.val)
                if cur.left != None:
                    queue.append(cur.left)
                if cur.right != None:
                    queue.append(cur.right)
            res.append(level)
        return res