Binary Tree Level Order Traversal
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? > read more on how binary tree is serialized on OJ.
Analyse:
1. Recursion: If the level does not exist, create it and then push corresponding value into it.
Runtime: 4ms.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int>> levelOrder(TreeNode* root) { 13 vector<vector<int> >result; 14 traverse(root, 0, result); 15 return result; 16 } 17 void traverse(TreeNode* root, int level, vector<vector<int> >& result){ 18 if(!root) return; 19 if(level == result.size()) //the level does not exist and need to create it 20 result.push_back(vector<int> ()); 21 22 result[level].push_back(root->val); 23 traverse(root->left, level + 1, result); 24 traverse(root->right, level + 1, result); 25 } 26 };
2. Iteration: Using queue to store nodes. When poping the first node, we need to add its children(child) to the queue. Then keep poping until the queue is empty.
Runtime: 8ms.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int> > levelOrder(TreeNode* root) { 13 vector<vector<int> > result; 14 if(!root) return result; 15 16 queue<TreeNode* > qu; 17 qu.push(root); 18 while(!qu.empty()){ 19 int n = qu.size(); 20 vector<int> level; //store the visited nodes in current level 21 while(n--){ 22 TreeNode* temp = qu.front(); //pop the first node in the queue 23 level.push_back(temp->val); 24 qu.pop(); 25 if(temp->left) qu.push(temp->left); //add its children(child) 26 if(temp->right) qu.push(temp->right); 27 } 28 result.push_back(level); 29 } 30 return result; 31 } 32 };
