leetcode笔记:Binary Tree Level Order Traversal
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一. 题目描述
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,#,#,15,7},
3 / \ 9 20 / \ 15 7
二. 题目分析
无。
三. 示例代码
参考了Discussion中stellari的做法,递归进行层次遍历,并将每个level对应于相应的vector:
#include <iostream>#include <vector>using namespace std;struct TreeNode{ int val; TreeNode *left; TreeNode *right; TreeNode(int x) :val(x), left(NULL), right(NULL){}};class Solution {public: vector<vector<int> > levelOrder(TreeNode *root) { vector<vector<int> > result; orderTraversal(root, 1, result); return result; }private: void orderTraversal(TreeNode* root, int level, vector<vector<int> > & result) { if (root == NULL) return; if (level > result.size()) // 每往下一级,创建一个空的vector保存该层数据 result.push_back(vector<int>()); result[level - 1].push_back(root->val); orderTraversal(root->left, level + 1, result); orderTraversal(root->right, level + 1, result); }};
四. 小结
以上为递归实现,也可以写成迭代版本,时间复杂度O(n),空间复杂度O(1)。
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