Binary Tree Level Order Traversal LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Binary Tree Level Order Traversal Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[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].
  • -1000 <= Node.val <= 1000

Binary Tree Level Order Traversal Solutions

Time: O(n)
Space: O(n)

C++

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

    vector<vector<int>> ans;
    queue<TreeNode*> q{{root}};

    while (!q.empty()) {
      vector<int> currLevel;
      for (int sz = q.size(); sz > 0; --sz) {
        TreeNode* node = q.front();
        q.pop();
        currLevel.push_back(node->val);
        if (node->left)
          q.push(node->left);
        if (node->right)
          q.push(node->right);
      }
      ans.push_back(currLevel);
    }

    return ans;
  }
};

Java

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

    List<List<Integer>> ans = new ArrayList<>();
    Queue<TreeNode> q = new ArrayDeque<>(Arrays.asList(root));

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

    return ans;
  }
}

Python

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

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

    while q:
      currLevel = []
      for _ in range(len(q)):
        node = q.popleft()
        currLevel.append(node.val)
        if node.left:
          q.append(node.left)
        if node.right:
          q.append(node.right)
      ans.append(currLevel)

    return ans

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