Leetcode Note: Binary Tree level Order traversal

I. Title 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 {3,9,20,#,#,15,7},

   9 20     15  7

Two. Topic analysis

No.

Three. Sample code

Referring to the practice of Stellari in discussion, recursive hierarchy traversal, and each level corresponds to the corresponding vector:

#include <iostream>#include <vector>using namespace STD;structtreenode{intVal    TreeNode *left;    TreeNode *right; TreeNode (intx): Val (x), left (null), right (null) {}};classSolution { Public: vector<vector<int> >Levelorder (TreeNode *root) { vector<vector<int> >Result Ordertraversal (Root,1, result);returnResult }Private:voidOrdertraversal (treenode* root,intLevel vector<vector<int> >& result) {if(Root = NULL)return;if(Level > Result.size ())//Each down level, create an empty vector to save the layer dataResult.push_back ( vector<int>()); Result[level-1].push_back (Root->val); Ordertraversal (Root->left, Level +1, result); Ordertraversal (Root->right, Level +1, result); }};

Four. Summary

The above is recursive implementation, can also be written as iteration version, time complexity O (n), Space complexity O (1).

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Leetcode Note: Binary Tree level Order traversal