Binary Tree Zigzag Level Order Traversal
Problem Statement - link #
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).
Examples #
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 #
-100 <= Node.val <= 100- The number of nodes in the tree is in the range
[0, 2000].
Solutions #
/**
* 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;
bool even = true;
q.push(root);
while(q.size()) {
int n = q.size();
vector<int> temp;
while(n--) {
TreeNode* cur = q.front(); q.pop();
temp.push_back(cur->val);
if(cur->left) {
q.push(cur->left);
}
if(cur->right) {
q.push(cur->right);
}
}
if(!even) {
reverse(temp.begin(), temp.end());
}
res.push_back(temp);
even = !even;
}
return res;
}
};