算法：BFS和DFS相关题目

最新推荐文章于 2025-02-12 14:51:29 发布

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最新推荐文章于 2025-02-12 14:51:29 发布 · 378 阅读

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#leetcode #算法 #java

BFS和DFS相关题目

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BFS和DFS相关题目

一、102. 二叉树的层序遍历

1.1 解法一：BFS

    public List<List<Integer>> levelOrder(TreeNode root) {
   
   
        if (root==null) return new ArrayList<>();
        List<List<Integer>> result = new ArrayList<>();
        Deque<TreeNode> deque = new LinkedList<>();
        deque.addLast(root);
        while (!deque.isEmpty()) {
   
   
            int n = deque.size();
            List<Integer> list = new ArrayList<>(n);
            for (int i=0; i<n; i++) {
   
   
                TreeNode node = deque.pop();
                list.add(node.val);
                if (node.left!=null) deque.addLast(node.left);
                if (node.right!=null) deque.addLast(node.right);
            }
            result.add(list);
        }
        return result;
    }

二、433. 最小基因变化

2.1 解法一：BFS

    public int minMutation(String start, String end, String[] bank) {
   
   
        Set<String> bankSet = new HashSet<>(Arrays.asList(bank));
        char[] dir = {
   
   'A', 'C', 'G', 'T'};
        bankSet.remove(start);
        Set<String> visited = new HashSet<>();
        Deque<String> deque = new LinkedList<>();
        deque.addLast(start);
        visited.add(start);
        int res = 0;
        while (!deque.isEmpty()) {
   
   
            int n = deque.size();
            for (int i=0; i<n; i++) {
   
   
                String str = deque.pop();
                if (str.equals(end)) return res;
                char[] a = str.toCharArray();
                for (int j=0; j<a.length; j++) {
   
   
                    char old = a[j];
                    for (char c: dir) {
   
   
                        if (old==c) continue;
                        a[j] = c;
                        String s = String.valueOf(a);
                        if (bankSet.contains(s) && !visited.contains(s)) {
   
   
                            visited.add(s);
                            deque.addLast(s);
                        }
                        a[j] = old;
                    }
                }
            }
            res++;
        }
        return -1;
    }

三、22. 括号生成

3.1 解法一：BFS + 验证

    public List<String> generateParenthesis(int n) {
        List<String> result =new ArrayList<>();
        Deque<String> deque = new LinkedList<>();
        deque.addLast("");
        int level = 0, h = n * 2;
        while (level<h) {
            int size = deque.size();
            for (int i=0; i<size; i++) {
                String s = deque.pop();
                deque.addLast(s+"(");
                deque.addLast(s+")");
            }
            level++;
        }
        while(!deque.isEmpty()) {
            String s = deque.pop();
            if (valid(s))