Leetcode Note: Binary Tree Level Order Traversal

Leetcode Note: Binary Tree Level Order Traversal

I. Description

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 {, 20, #, #, 15, 7 },

  3 /  9 20   /   15  7

Ii. Question Analysis

None.

Iii. Sample Code

Based on the stellari practice in Discussion, recursively traverse layers and map each level to the corresponding vector:

# Include
  
   
# Include
   
    
Using namespace std; struct TreeNode {int val; TreeNode * left; TreeNode * right; TreeNode (int x): val (x), left (NULL), right (NULL) {}}; class Solution {public: vector
    
     
> LevelOrder (TreeNode * root) {vector
     
      
> Result; orderTraversal (root, 1, result); return result;} private: void orderTraversal (TreeNode * root, int level, vector
      
        > & Result) {if (root = NULL) return; if (level> result. size () // create an empty vector next to each level to save the result of this layer of data. push_back (vector
       
         (); Result [level-1]. push_back (root-> val); orderTraversal (root-> left, level + 1, result); orderTraversal (root-> right, level + 1, result );}};
       
      
     
    
   
  

Iv. Summary

The above is a recursive implementation. It can also be written as an iterative version, with time complexity O (n) and space complexity O (1 ).