g0101_0200.s0111_minimum_depth_of_binary_tree.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0111_minimum_depth_of_binary_tree;

// #Easy #Depth_First_Search #Breadth_First_Search #Tree #Binary_Tree #Udemy_Tree_Stack_Queue
// #2022_06_23_Time_1_ms_(97.49%)_Space_97_MB_(5.22%)

import com_github_leetcode.TreeNode;
import java.util.LinkedList;
import java.util.Queue;

/*
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
/**
 * 111 - Minimum Depth of 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/12/ex_depth.jpg)
 *
 * **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`
**/
public class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        Queue queue = new LinkedList<>();
        queue.add(root);
        int d = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode current = queue.poll();
                if (current.left == null && current.right == null && d > 0) {
                    return d + 1;
                }
                if (current.right != null) {
                    queue.add(current.right);
                }
                if (current.left != null) {
                    queue.add(current.left);
                }
            }
            d++;
        }
        return d;
    }
}