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题目描述:(链接)
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.
解题思路：
广度优先遍历
递归版：
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 travel(root, 1); 14 return result; 15 } 16 17 void travel(TreeNode *root, int level) { 18 if (!root) return; 19 if (level > result.size()) { 20 result.push_back(vector<int>()); 21 } 22 23 result[level - 1].push_back(root->val); 24 travel(root->left, level + 1); 25 travel(root->right, level + 1); 26 } 27 private: 28 vector<vector<int>> result; 29 };
迭代版:
借助一个队列实现先进先出：
/** * Definition for a binary tree node. * 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; if (!root) return result; queue<TreeNode *> current, next; vector<int> level; current.push(root); while (!current.empty()) { while (!current.empty()) { TreeNode *tmp = current.front(); current.pop(); level.push_back(tmp->val); if (tmp->left != nullptr) next.push(tmp->left); if (tmp->right != nullptr) next.push(tmp->right); } result.push_back(level); level.clear(); swap(current, next); } return result; } };
[LeetCode]Binary Tree Level Order Traversal
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原文地址：http://www.cnblogs.com/skycore/p/5004692.html
