103. Binary Tree Zigzag Level Order Traversal

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103. Binary Tree Zigzag Level Order Traversal

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:

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Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]

Example 2:

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Input: root = [1]
Output: [[1]]

Example 3:

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Input: root = []
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 2000].
  • -100 <= Node.val <= 100

Hints/Notes

Solution

Language: C++

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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 level = 1;
        while (!q.empty()) {
            int size = q.size();
            vector<int> cur;
            for (int i = 0; i < size; i++) {
                TreeNode* node = q.front();
                q.pop();
                cur.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            if (level % 2 == 0) ranges::reverse(cur);
            res.push_back(cur);
            level++;
        }
        return res;
    }
};