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Binary Tree Zigzag Level Order Traversal⚓︎

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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

Solution⚓︎

Way 1⚓︎

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/**
 * 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 cnt = 0;
        while (!q.empty()) {
            vector<int> level;
            int len = q.size();

            while (len--) {
                auto node = q.front();
                q.pop();
                level.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }

            if (++cnt % 2 == 0) reverse(level.begin(), level.end());
            res.push_back(level);
        }
        return res;
    }
};

Way 2⚓︎

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/**
 * 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 isOrderLeft = true;

        while (!q.empty()) {
            int len = q.size();
            deque<int> level;

            while (len--) {
                TreeNode* node = q.front();
                q.pop();

                if (isOrderLeft) level.push_back(node->val);
                else level.push_front(node->val);

                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            res.emplace_back(vector<int>(level.begin(), level.end()));
            isOrderLeft = !isOrderLeft;
        }
        return res;
    }
};
  • Time complexity: \(O(N)\);
  • Space complexity: \(O(N)\).