7.5 Minimum Depth of a Binary Tree (easy)

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

import java.util.*;

class TreeNode {
  int val;
  TreeNode left;
  TreeNode right;

  TreeNode(int x) {
    val = x;
  }
};

class Main {
  public static int findDepth(TreeNode root) {
    if(root == null) return 0;
    Queue<TreeNode> queue = new LinkedList<>();
    queue.offer(root);
    int depth = 1;
    while(!queue.isEmpty()) {
      int levelSize = queue.size();
      for(int i = 0; i < levelSize; i++) {
        TreeNode currNode = queue.poll();

        if(currNode.left != null) queue.offer(currNode.left);
        if(currNode.right != null) queue.offer(currNode.right);
        if(currNode.left == null && currNode.right == null) {
          return depth;
        }

      }
      depth++;
    }
    return depth;
  }

  public static void main(String[] args) {
    TreeNode root = new TreeNode(12);
    root.left = new TreeNode(7);
    root.right = new TreeNode(1);
    root.right.left = new TreeNode(10);
    root.right.right = new TreeNode(5);
    System.out.println("Tree Minimum Depth: " + Main.findDepth(root));
    root.left.left = new TreeNode(9);
    root.right.left.left = new TreeNode(11);
    System.out.println("Tree Minimum Depth: " + Main.findDepth(root));
  }
}

Time complexity #

The time complexity of the above algorithm is O(N), where ‘N’ is the total number of nodes in the tree. This is due to the fact that we traverse each node once.

Space complexity #

The space complexity of the above algorithm will be O(N) which is required for the queue. Since we can have a maximum of N/2 nodes at any level (this could happen only at the lowest level), therefore we will need O(N) space to store them in the queue