Leetcode

Minimum Depth of Binary Tree

May 26, 2022 by aAshish

Approach 1: Top-down DFS

  • Time:O(n)
  • Space:O(h)

C++

 
class Solution {
 public:
  int minDepth(TreeNode* root) {
    if (!root)
      return 0;
    if (!root->left)
      return minDepth(root->right) + 1;
    if (!root->right)
      return minDepth(root->left) + 1;
    return min(minDepth(root->left), minDepth(root->right)) + 1;
  }
};

JAVA

 
class Solution {
  public int minDepth(TreeNode root) {
    if (root == null)
      return 0;
    if (root.left == null)
      return minDepth(root.right) + 1;
    if (root.right == null)
      return minDepth(root.left) + 1;
    return Math.min(minDepth(root.left), minDepth(root.right)) + 1;
  }
}

Python

 
class Solution:
  def minDepth(self, root: Optional[TreeNode]) -> int:
    if not root:
      return 0
    if not root.left:
      return self.minDepth(root.right) + 1
    if not root.right:
      return self.minDepth(root.left) + 1
    return min(self.minDepth(root.left), self.minDepth(root.right)) + 1

Approach 2: Bottom-up BFS

  • Time:O(n)
  • Space:O(n)

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)