Minimum Depth of Binary Tree LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Minimum Depth of Binary Tree Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 2

Example 2:

Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5

Constraints:

  • The number of nodes in the tree is in the range [0, 105].
  • -1000 <= Node.val <= 1000

Minimum Depth of Binary Tree Solutions

Time: O(n)
Space: O(h)

C++

class Solution {
 public:
  int minDepth(TreeNode* root) {
    if (!root)
      return 0;

    int ans = 0;
    queue<TreeNode*> q{{root}};

    while (!q.empty()) {
      ++ans;
      for (int sz = q.size(); sz > 0; --sz) {
        TreeNode* node = q.front();
        q.pop();
        if (!node->left && !node->right)
          return ans;
        if (node->left)
          q.push(node->left);
        if (node->right)
          q.push(node->right);
      }
    }

    throw;
  }
};

Java

 class Solution {
  public int minDepth(TreeNode root) {
    if (root == null)
      return 0;

    int ans = 0;
    Queue<TreeNode> q = new ArrayDeque<>(Arrays.asList(root));

    while (!q.isEmpty()) {
      ++ans;
      for (int sz = q.size(); sz > 0; --sz) {
        TreeNode node = q.poll();
        if (node.left == null && node.right == null)
          return ans;
        if (node.left != null)
          q.offer(node.left);
        if (node.right != null)
          q.offer(node.right);
      }
    }

    throw new IllegalArgumentException();
  }
}

Python


class Solution:
  def minDepth(self, root: Optional[TreeNode]) -> int:
    if not root:
      return 0

    ans = 0
    q = deque([root])

    while q:
      ans += 1
      for _ in range(len(q)):
        node = q.popleft()
        if not node.left and not node.right:
          return ans
        if node.left:
          q.append(node.left)
        if node.right:
          q.append(node.right)

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