[Simple Algorithm] 29. Hierarchical Traversal of Binary Tree

topic:

Given a binary tree, return its node values ​​traversed hierarchically. (i.e. layer by layer, visiting all nodes from left to right).

E.g:
Given a binary tree: [ 3 , 9 , 20 , null , null , 15 , 7 ],

    3
   / \
  9  20
    /  \
   15   7
Return its hierarchical traversal result:

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

Problem solving ideas:

The queue can be used, and the nodes of each layer can be put into the queue in sequence according to the level.

/**
 * 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 {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        queue<TreeNode *> qu;
        
        if(root == NULL){
            return res;
        }
        
        /*level travel*/
        qu.push(root);    
        while(!qu.empty()){
            int length = qu.size();
            vector<int> level;
            for(int i = 0;i < length;++i){
                TreeNode * node = qu.front();
                qu.pop();
                level.push_back(node->val);
                if(node->left){
                    qu.push(node->left);
                }
                if(node->right){
                    qu.push(node->right);
                }
            }
            res.push_back(level);
        }
        
        return res;
    }
};