Binary Tree Zigzag Level Order Traversal LeetCode Solution

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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

Binary Tree Zigzag Level Order Traversal Solutions

Time: O(n)
Space: O(n)

C++

class Solution {
 public:
  vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
    if (!root)
      return {};

    vector<vector<int>> ans;
    deque<TreeNode*> q{{root}};
    bool isLeftToRight = true;

    while (!q.empty()) {
      vector<int> currLevel;
      for (int sz = q.size(); sz > 0; --sz)
        if (isLeftToRight) {
          TreeNode* node = q.front();
          q.pop_front();
          currLevel.push_back(node->val);
          if (node->left)
            q.push_back(node->left);
          if (node->right)
            q.push_back(node->right);
        } else {
          TreeNode* node = q.back();
          q.pop_back();
          currLevel.push_back(node->val);
          if (node->right)
            q.push_front(node->right);
          if (node->left)
            q.push_front(node->left);
        }
      ans.push_back(currLevel);
      isLeftToRight = !isLeftToRight;
    }

    return ans;
  }
};

Java

 
class Solution {
  public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
    if (root == null)
      return new ArrayList<>();

    List<List<Integer>> ans = new ArrayList<>();
    Deque<TreeNode> q = new ArrayDeque<>(Arrays.asList(root));
    boolean isLeftToRight = true;

    while (!q.isEmpty()) {
      List<Integer> currLevel = new ArrayList<>();
      for (int sz = q.size(); sz > 0; --sz)
        if (isLeftToRight) {
          TreeNode node = q.pollFirst();
          currLevel.add(node.val);
          if (node.left != null)
            q.addLast(node.left);
          if (node.right != null)
            q.addLast(node.right);
        } else {
          TreeNode node = q.pollLast();
          currLevel.add(node.val);
          if (node.right != null)
            q.addFirst(node.right);
          if (node.left != null)
            q.addFirst(node.left);
        }
      ans.add(currLevel);
      isLeftToRight = !isLeftToRight;
    }

    return ans;
  }
}

Python


class Solution:
  def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
    if not root:
      return []

    ans = []
    q = deque([root])
    isLeftToRight = True

    while q:
      currLevel = []
      for _ in range(len(q)):
        if isLeftToRight:
          node = q.popleft()
          currLevel.append(node.val)
          if node.left:
            q.append(node.left)
          if node.right:
            q.append(node.right)
        else:
          node = q.pop()
          currLevel.append(node.val)
          if node.right:
            q.appendleft(node.right)
          if node.left:
            q.appendleft(node.left)
      ans.append(currLevel)
      isLeftToRight = not isLeftToRight

    return ans

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