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] ]
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
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//递归方法:用一个bool记录是从左到右还是从右到左,每一层结束就翻转一下。 class Solution { public: vector<vector<int>> zigzagLevelOrder(TreeNode* root) { vector<vector<int>> res; traverse(root, 1, res, true); return res; } void traverse(TreeNode* root, int level, vector<vector<int>> &res, bool left_to_right){ if (!root) return; if (level > res.size()) res.push_back(vector<int>()); if (left_to_right) res[level - 1].push_back(root->val); else res[level - 1].insert(res[level - 1].begin(), root->val); traverse(root->left, level + 1, res, !left_to_right); traverse(root->right, level + 1, res, !left_to_right); } };