Binary Tree Zigzag Level Order Traversal LeetCode Solution

Problem – Binary Tree Zigzag Level Order Traversal LeetCode Solution

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

Binary Tree Zigzag Level Order Traversal LeetCode Solution in Java

public class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) 
    {
        List<List<Integer>> sol = new ArrayList<>();
        travel(root, sol, 0);
        return sol;
    }
    
    private void travel(TreeNode curr, List<List<Integer>> sol, int level)
    {
        if(curr == null) return;
        
        if(sol.size() <= level)
        {
            List<Integer> newLevel = new LinkedList<>();
            sol.add(newLevel);
        }
        
        List<Integer> collection  = sol.get(level);
        if(level % 2 == 0) collection.add(curr.val);
        else collection.add(0, curr.val);
        
        travel(curr.left, sol, level + 1);
        travel(curr.right, sol, level + 1);
    }
}

Binary Tree Zigzag Level Order Traversal LeetCode Solution in C++

vector<vector<int> > zigzagLevelOrder(TreeNode* root) {
    if (root == NULL) {
        return vector<vector<int> > ();
    }
    vector<vector<int> > result;

    queue<TreeNode*> nodesQueue;
    nodesQueue.push(root);
    bool leftToRight = true;

    while ( !nodesQueue.empty()) {
        int size = nodesQueue.size();
        vector<int> row(size);
        for (int i = 0; i < size; i++) {
            TreeNode* node = nodesQueue.front();
            nodesQueue.pop();

            // find position to fill node's value
            int index = (leftToRight) ? i : (size - 1 - i);

            row[index] = node->val;
            if (node->left) {
                nodesQueue.push(node->left);
            }
            if (node->right) {
                nodesQueue.push(node->right);
            }
        }
        // after this level
        leftToRight = !leftToRight;
        result.push_back(row);
    }
    return result;
}

Binary Tree Zigzag Level Order Traversal LeetCode Solution in Python

class Solution:
    def zigzagLevelOrder(self, root):
        if not root: return []
        queue = deque([root])
        result, direction = [], 1
        
        while queue:
            level = []
            for i in range(len(queue)):
                node = queue.popleft()
                level.append(node.val)
                if node.left:  queue.append(node.left)
                if node.right: queue.append(node.right)
            result.append(level[::direction])
            direction *= (-1)
        return result

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