Description
Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: [[3],[20,9],[15,7]]
Example 2:
Input: root = [1] Output: 1
Example 3:
Input: root = [] Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -100 <= Node.val <= 100
Code
和 Binary Tree Level Order Traversal 類似，只是要考慮方向性。
Level order traversal with reversion
/**
* Definition for a binary tree node.
* 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) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> res;
if(!root) return res;
queue<TreeNode*> q;
q.push(root);
int level_n = 0;
while(!q.empty()) {
int levelSize = q.size();
vector<int> level;
for(int i = 0; i < levelSize; i++) {
auto node = q.front();
level.push_back(node->val);
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
q.pop();
}
if(level_n % 2 != 0) reverse(level.begin(), level.end());
res.push_back(level);
level_n++;
}
return res;
}
};# Level order traversal without reversion
關鍵在於：int index = reverse? levelSize - i - 1 : i;
/**
* Definition for a binary tree node.
* 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) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> res;
if(!root) return res;
queue<TreeNode*> q;
q.push(root);
bool reverse = false;
while(!q.empty()) {
int levelSize = q.size();
vector<int> level(levelSize);
for(int i = 0; i < levelSize; i++) {
auto node = q.front();
int index = reverse? levelSize - i - 1 : i;
level[index] = node->val;
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
q.pop();
}
reverse = !reverse;
res.push_back(level);
}
return res;
}
};