0103. Binary Tree Zigzag Level Order Traversal

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

Analysis

When seeing level order traversal, we should think about BFS. Different from level order traversal problem, it also requires us to reverse the level as we proceed, so we can introduce a bool right variable to represent if we need to reverse the order.

Code

/**
 * 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{{root}};
        bool right = true;
        while (!q.empty()) {
            int sz = q.size();
            vector<int> level;
            for (int i = 0; i < sz; ++i) {
                TreeNode* t = q.front();
                level.push_back(t -> val);
                q.pop();
                if (t -> left) q.push(t -> left);
                if (t -> right) q.push(t -> right);
            }
            if (right) res.push_back(level);
            else {
                reverse(level.begin(), level.end());
                res.push_back(level);
            }
            right = !right;
        }
        return res;
    }
};

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Last update: April 1, 2022