Min 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.

         1
        /
       2

min depth = 2.

Method 1:

Level order traversal/BFS

Solution:

Time: O(n)
Space: O(n)

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) {
 *      val = x;
 *      left=null;
 *      right=null;
 *     }
 * }
 */
public class Solution {
    public int minDepth(TreeNode A) {
        Deque<TreeNode> queue = new ArrayDeque<>();
        int min = 0;
        TreeNode curr = A;
        queue.add(curr);
        while (!queue.isEmpty()) {
            min ++;
            int size = queue.size();
            for (int i = 0; i < size; i ++) {
                TreeNode node = queue.poll();
                if (node.left == null && node.right == null) {
                    return min;
                }
                if (node.left != null) {
                    queue.add(node.left);
                } 
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
        }
        return min;
    }
}

Method 2:

DFS

Solution:

Time: O(n)
Space: O(1)

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) {
 *      val = x;
 *      left=null;
 *      right=null;
 *     }
 * }
 */
public class Solution {
    public int minDepth(TreeNode A) {
        if (A == null) {
            return 0;
        }
        if (A.left == null && A.right == null) {
            return 1;
        }
        if (A.left == null) {
            return minDepth(A.right) + 1;
        }
        if (A.right == null) {
            return minDepth(A.left) + 1;
        }
        int left = minDepth(A.left);
        int right = minDepth(A.right);
        return Math.min(left, right) + 1;
    }
}