[牛客网-Leetcode] #树简binary-tree-level-order-traversal

Binary tree level traversal binary-tree-level-order-traversal

Title description

Given a binary tree, return the result of the traversal of the binary tree sequence, (traversing layer by layer from left to right)

For example:
the given binary tree is {3,9,20,#,#,15,7},
3↵ / ↵ 9 20↵ / ↵ 15 7
The result of this binary tree sequence traversal is
[↵ [3],↵ [ 9,20],↵ [15,7]↵]

If you don’t know the meaning of "{1,#,2,3}", please continue reading.
We use the following method to serialize the
binary tree : the serialization of the binary tree follows the principle of layer order traversal, "#" means that the position is a line At the end of the path, there are no more nodes below.
For example:
1↵ / ↵ 2 3↵ /↵ 4↵ ↵ 5
The result of the above binary tree serialization is: "{1,2,3,#,#,4,#,#,5}".

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]↵]↵

confused what"{1,#,2,3}"means?

Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

1↵ / ↵ 2 3↵ /↵ 4↵ ↵ 5↵
The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

Example

Example 1
input

{1,2}

Output

[[1],[2]]

Example 2
input

{1,2,3,4,#,#,5}

Output

[[1],[2,3],[4,5]]

Problem-solving ideas

• The level traversal of the binary tree is solved by a queue. Pay attention to the number of nodes at each level

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
    
    
public:
    vector<vector<int> > levelOrder(TreeNode* root) {
    
    
        vector<vector<int> > res;
        if(root == NULL) return res;
        queue<TreeNode*> myque;
        myque.push(root);
        //只要队列不空
        while(!myque.empty()) {
    
    
            vector<int> cur;  //保存当前层的所有节点值
            int size = myque.size();  //每次先保存当前层的节点数
            //当前层的所有节点全部依次出队并加入cur，然后把下一层的节点全部放进来，等待下一次大循环
            for(int i = 0; i < size; i ++) {
    
    
                TreeNode* p = myque.front();
                myque.pop();
                cur.push_back(p -> val);
                if(p -> left) myque.push(p -> left);
                if(p -> right) myque.push(p -> right);
            }
            res.push_back(cur);
        }
        return res;
    }
};