Breadth-first Search -- Leetcode problem102. Binary Tree Level Order Traversal
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- 描述: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,null,null,15,7],
3 / \ 9 20 / \ 15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
- 分析:层序遍历并输出每层的值
- 思路一:bfs层序遍历
class Solution {public: vector<vector<int>> levelOrder(TreeNode* root) { vector<vector<int>> my_vec; vector<int> result; if (!root) return my_vec; queue<TreeNode*> my_queue; my_queue.push(root); while (!my_queue.empty()) { int n = my_queue.size(); result.clear(); for (int i = 0; i < n; i ++) { TreeNode* temp = my_queue.front(); my_queue.pop(); result.push_back(temp -> val); if (temp -> left) my_queue.push(temp -> left); if (temp -> right) my_queue.push(temp -> right); } my_vec.push_back(result); } return my_vec;}};
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