# 二叉树的锯齿形层序遍历

给定一个二叉树,返回其节点值的锯齿形层序遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。

例如:
给定二叉树 [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回锯齿形层序遍历如下:

[
  [3],
  [20,9],
  [15,7]
]

以下错误的选项是?

## aop ### before ```c #include using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode() : val(0), left(nullptr), right(nullptr) {} TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} }; ``` ### after ```c ``` ## 答案 ```c class Solution { public: vector> zigzagLevelOrder(TreeNode *root) { vector> result; if (root == NULL) { return result; } deque node_queue; node_queue.push_back(root); node_queue.push_back(NULL); deque level_list; bool is_order_left = true; while (node_queue.size() > 0) { TreeNode *curr_node = node_queue.front(); node_queue.pop_front(); if (curr_node != NULL) { if (curr_node->left != NULL) { node_queue.push_back(curr_node->left); } if (curr_node->right != NULL) { node_queue.push_back(curr_node->right); } } else { vector tmp; for (int a : level_list) { tmp.push_back(a); } result.push_back(tmp); level_list.clear(); if (node_queue.size() > 0) { node_queue.push_back(NULL); } is_order_left = !is_order_left; } } return result; } }; ``` ## 选项 ### A ```c class Solution { public: vector> zigzagLevelOrder(TreeNode *root) { if (root == NULL) return {}; vector> res; int flag = 1; stack a, b; a.push(root); while (!a.empty() || !b.empty()) { vector cur; while (flag == 1 && !a.empty()) { root = a.top(); cur.push_back(root->val); a.pop(); if (root->left != NULL) b.push(root->left); if (root->right != NULL) b.push(root->right); } while (flag == -1 && !b.empty()) { root = b.top(); cur.push_back(root->val); b.pop(); if (root->right != NULL) a.push(root->right); if (root->left != NULL) a.push(root->left); } flag = -1 * flag; res.push_back(cur); } return res; } }; ``` ### B ```c class Solution { public: vector> zigzagLevelOrder(TreeNode *root) { vector> ans; if (!root) return ans; queue qnode; bool orderByLeft = true; qnode.push(root); while (!qnode.empty()) { int levelSize = qnode.size(); deque level; for (int i = 0; i < levelSize; i++) { auto curNode = qnode.front(); qnode.pop(); if (orderByLeft) { level.push_back(curNode->val); } else { level.push_front(curNode->val); } if (curNode->left != NULL) qnode.push(curNode->left); if (curNode->right != NULL) qnode.push(curNode->right); } orderByLeft = !orderByLeft; vector curlevel{level.begin(), level.end()}; ans.push_back(curlevel); } return ans; } }; ``` ### C ```c class Solution { public: vector> zigzagLevelOrder(TreeNode *root) { if (!root) return {}; vector> res; queue q{{root}}; int cnt = 0; while (!q.empty()) { vector oneLevel; for (int i = q.size(); i > 0; i--) { TreeNode *t = q.front(); q.pop(); oneLevel.push_back(t->val); if (t->left) q.push(t->left); if (t->right) q.push(t->right); } if (cnt % 2 == 1) reverse(oneLevel.begin(), oneLevel.end()); res.push_back(oneLevel); cnt++; } return res; } }; ``` ### D ```c class Solution { public: vector> zigzagLevelOrder(TreeNode *root) { vector> results; if (root == NULL) { return results; } zigzagLevelOrderDFS_helper(root, 0, results); return results; } void zigzagLevelOrderDFS_helper(TreeNode *node, int level, vector> &results) { if (level >= results.size()) { vector new_level; new_level.push_back(node->val); results.push_back(new_level); } else { if (level % 2 == 0) { results[level].push_back(node->val); } else { results[level].insert(results[level].begin(), node->val); } } if (node->left != NULL) { zigzagLevelOrderDFS_helper(node->left, level + 1, results); } if (node->right != NULL) { zigzagLevelOrderDFS_helper(node->right, level + 1, results); } } }; ```