Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree {3,9,20,#,#,15,7}
,
3 / \ 9 20 / \ 15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
此题所谓的Zig-Zag 遍历,及第一行从左到有,第二行从右到左,第三行从左到右,依次类推。此题仍然可以在层次遍历后,对结果进行处理。代码如下:
class Solution { public: vector<vector<int>> zigzagLevelOrder(TreeNode* root) { //同层次遍历 vector<vector<int>>result; if (!root) return result; queue<TreeNode*> vec; vec.push(root); while (vec.size() > 0) { queue<TreeNode*> tmp_vec; vector<int> tmp_result; while (vec.size() > 0) { TreeNode* node = vec.front(); vec.pop(); if (node->left) tmp_vec.push(node->left); if (node->right) tmp_vec.push(node->right); tmp_result.push_back(node->val); } vec = tmp_vec; result.push_back(tmp_result); } //对第2层、4层。。。进行翻转 for (int i = 1; i < result.size(); i += 2) { reverse(result[i].begin(), result[i].end()); } return result; } };