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[LeetCode] 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,#,
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 vector<vector<int> > result;14 if (root == nullptr) return result;15 16 queue<TreeNode *> current, next;17 vector<int> level;18 19 current.push(root);20 while (!current.empty()) {21 while (!current.empty()) {22 TreeNode *p = current.front();23 current.pop();24 level.push_back(p->val);25 if (p->left != nullptr) next.push(p->left);26 if (p->right != nullptr) next.push(p->right);27 }28 result.push_back(level);29 level.clear();30 swap(current, next);31 }32 33 return result;34 }35 };
#,15,7},
3 / 9 20 / 15 7
return its level order traversal as:
[ [3], [9,20], [15,7]]
思路：用两个vector分层次记录值，时间复杂度O(n)，空间复杂度O(n)
[LeetCode] Binary Tree Level Order Traversal
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