DSA

Binary Tree Zigzag Level Order Traversal

Binary Trees. Time O(n), Space O(n).

August 8, 2026

DSA Binary Tree LeetCode Practice Arrays BFS Matrix

Practice here

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).

Implementation (BFS - Breadth-First Search)#

Intuition: Standard BFS processes nodes left-to-right on every level. To produce a zigzag, you only need to reverse the collected values of every odd-numbered level — the traversal order itself stays the same, only the output array is flipped.
Mechanics: A levelId counter tracks the current level. All nodes at the current level are dequeued and their values pushed into a temporary vector. If levelId is odd, the vector is reversed before being added to the result. Children are always enqueued left-then-right regardless of direction.
Trade-off: Reversing after collection is O(w) per level (w = level width) and adds negligible cost relative to the O(n) BFS. An alternative is to use a deque and push from the front or back depending on direction, avoiding explicit reversal — both strategies are O(n) overall.

Visit all nodes at level 0 → then level 1 → then level 2 → …

Perform normal level-order traversal using a queue, but reverse the order of values every alternate level to simulate the zig-zag pattern.

cpp

/**
 * 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) {
        if(!root)
            return {};

        vector<vector<int>> traversalResult;
        queue<TreeNode*> q;
        q.push(root);

        int levelId = 0;

        while(!q.empty())
        {
            int size = q.size();
            vector<int> levelResult;
            for(int i=0;i<size;i++)
            {
                TreeNode* currNode = q.front();
                q.pop();

                levelResult.push_back(currNode->val);

                if(currNode->left)
                    q.push(currNode->left);

                if(currNode->right)
                    q.push(currNode->right);
            }
            if(levelId%2!=0){
                reverse(levelResult.begin(), levelResult.end());
                traversalResult.push_back(levelResult);
            }
            else
                traversalResult.push_back(levelResult);

            levelId++;
        }
        return traversalResult;
    }
};

Time Complexity: O(n), Each node visited once

Space Complexity: O(n), the queue can store all the nodes of the last level.

DSAShortest Path in Binary MatrixGraphs. Time O(V^2).DSARotate MatrixSimulation approach — transpose + row-reverse. Time O(n²), Space O(1).DSASpiral MatrixSimulation approach.

Implementation (BFS - Breadth-First Search)#

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