力扣算法 Java 刷题笔记【BFS 篇】hot100（一）打开转盘锁、二叉树的最小深度 2

力扣算法 Java 刷题笔记【BFS 篇】hot100（一）打开转盘锁、二叉树的最小深度 2

文章目录

1. 打开转盘锁（中等）2. 二叉树的最小深度（简单）

1. 打开转盘锁（中等）

地址: https://leetcode-cn.com/problems/open-the-lock/
2022/01/30
做题反思：

class Solution {
    public int openLock(String[] deadends, String target) {
        Set deads = new HashSet<>();
        for (String s : deadends) {
            deads.add(s);
        }
        Queue q = new linkedList<>();
        Set visited = new HashSet<>();
        q.offer("0000");
        visited.add("0000");
        int step = 0;

        while (!q.isEmpty()) {
            int sz = q.size();
            for (int i = 0; i < sz; i++) {
                String cur = q.poll();
                if (cur.equals(target)) {
                    return step;
                }
                if (deads.contains(cur)) {
                    continue;
                }
                for (int j = 0; j < 4; j++) {
                    String up = addOne(cur, j);
                    if (!visited.contains(up)) {
                        q.offer(up);
                        visited.add(up);
                    }

                    String down = minunesOne(cur, j);
                    if (!visited.contains(down)) {
                        q.offer(down);
                        visited.add(down);
                    }
                }
            }
            step++;
        }
        return -1;
    }

    String addOne(String cur, int j) {
        char[] ch = cur.toCharArray();
        if (ch[j] == '9') {
            ch[j] = '0';
        } else {
            ch[j]++;
        }
        return new String(ch);
    }

    String minunesOne(String cur, int j) {
        char[] ch = cur.toCharArray();
        if (ch[j] == '0') {
            ch[j] = '9';
        } else {
            ch[j]--;
        }
        return new String(ch);
    }
}

2. 二叉树的最小深度（简单）

地址: https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/
2022/01/28
做题反思：
递归

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

BFS

class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        Queue q = new linkedList<>();
        q.offer(root);
        int depth = 1;
        
        while(!q.isEmpty()) {
            int sz = q.size();
            for (int i = 0; i < sz; i++) {
                TreeNode cur = q.poll();
                if (cur.left == null && cur.right == null) {
                    return depth;
                }
                if (cur.left != null) {
                    q.offer(cur.left);
                }
                if (cur.right != null) {
                    q.offer(cur.right);
                }
            }
            depth++;
        }
        return depth;
    }
}