[Leetcode 77] 102 Binary Tree Level Order Traversal
Problem:
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] ]
Analysis:
The problem is an extension of the original level-order traversal problem. We need an extra sentinal node to signal the end of a level.
Each time we encounter it, we know that a level is ended so we push the partial_result_vector into the result vector and then immediately push it into the queue again.
The quit condition switch from queue.empty() to there's only the sentinal node in the queue.
Code:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 /** 2 * Definition for binary tree 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 // Start typing your C/C++ solution below 14 // DO NOT write int main() function 15 vector<vector<int> > res; 16 if (root == NULL) 17 return res; 18 19 queue<TreeNode *> q; 20 q.push(root); 21 22 TreeNode *tmp, *sentl; 23 24 sentl = new TreeNode((1<<31)); 25 sentl->left = sentl->right = NULL; 26 q.push(sentl); 27 28 vector<int> pres; 29 while (!(q.size()==1 && q.front() == sentl)) { 30 tmp = q.front(); 31 q.pop(); 32 33 if (tmp == sentl) { 34 res.push_back(pres); 35 pres.clear(); 36 q.push(sentl); 37 } else { 38 pres.push_back(tmp->val); 39 40 if (tmp->left != NULL) 41 q.push(tmp->left); 42 43 if (tmp->right != NULL) 44 q.push(tmp->right); 45 } 46 47 } 48 49 res.push_back(pres); 50 51 return res; 52 } 53 };