# 103. Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

Example:

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

# Solution

Approach 1: DFS.

Approach 2: BFS with queue.

Approach 3: BFS with two stacks.

# Code (Python)

Approach 1:

from collections import deque
class Solution:
    def zigzagLevelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        # recursive
        result = []
        self._traverse(root, 0, result)
        return list(map(list, result))
    
    def _traverse(self, node, level, result):
        if not node:
            return
        if len(result) == level:
            result.append(deque())
        if level % 2 == 0:
            result[level].append(node.val)
        else:
            result[level].appendleft(node.val)
        if node.left:
            self._traverse(node.left, level + 1, result)
        if node.right:
            self._traverse(node.right, level + 1, result)

Approach 2:

    def zigzagLevelOrder(self, root):
        # iterative
        if not root:
            return []
        result = []
        level = [root]
        reverse = False
        while level:
            current_level = deque()
            next_level = []
            for node in level:
                if not reverse:
                    current_level.append(node.val)
                else:
                    current_level.appendleft(node.val)
                if node.left:
                    next_level.append(node.left)
                if node.right:
                    next_level.append(node.right)
            result.append(list(current_level))
            level = next_level
            reverse = (not reverse)
        return result

# Code (C++)

Approach 1:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
// DFS.
class Solution {
private:
    vector<vector<int>> res;
    void doZigzagLevelOrder(TreeNode *root, int level) {
        if (root == NULL)
            return;
        if (res.size() < level)
            res.push_back(vector<int>());
        if (level % 2 != 0)
            res[level-1].push_back(root->val);
        else
            res[level-1].insert(res[level-1].begin(), root->val);
        doZigzagLevelOrder(root->left, level + 1);
        doZigzagLevelOrder(root->right, level + 1);
    }
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        doZigzagLevelOrder(root, 1);
        return res;
    }
};

Approach 2:

// BFS with queue.
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> res;
        queue<TreeNode*> q;
        if (root)
            q.push(root);
        bool leftToRight = true;
        while (!q.empty())
        {
            int qSize = q.size();
            vector<int> buf;
            for (int i = 0; i < qSize; ++i)
            {
                TreeNode *curr = q.front();
                q.pop();
                if (leftToRight)
                    buf.push_back(curr->val);
                else
                    buf.insert(buf.begin(), curr->val);
                if (curr->left)
                    q.push(curr->left);
                if (curr->right)
                    q.push(curr->right);
            }
            if (buf.size() > 0)
                res.push_back(buf);
            leftToRight = !leftToRight;
        }
        return res;
    }
};

Approach 3:

// BFS with two stacks.
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> res;
        stack<TreeNode*> st[2];
        int stIdx = 0;
        if (root)
            st[stIdx].push(root);
        while (!st[stIdx].empty())
        {
            vector<int> buf;
            while (!st[stIdx].empty())
            {
                TreeNode *curr = st[stIdx].top();
                st[stIdx].pop();
                buf.push_back(curr->val);
                if (stIdx == 0)
                {
                    if (curr->left)
                        st[1].push(curr->left);
                    if (curr->right)
                        st[1].push(curr->right);
                }
                else
                {
                    if (curr->right)
                        st[0].push(curr->right);
                    if (curr->left)
                        st[0].push(curr->left);
                }
            }
            if (buf.size() > 0)
                res.push_back(buf);
            stIdx = 1 - stIdx;
        }
        return res;
    }
};