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Java C++ 寻找重复子树

AnjaVon 人气:0

题目要求

思路一:DFS+序列化

Java

class Solution {
    Map<String, Integer> map = new HashMap<>();
    List<TreeNode> res = new ArrayList<>();
    public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
        DFS(root);
        return res;
    }
    String DFS(TreeNode root) {
        if (root == null)
            return " ";
        StringBuilder sb = new StringBuilder();
        sb.append(root.val).append("-");
        sb.append(DFS(root.left)).append(DFS(root.right));
        String sub = sb.toString(); // 当前子树
        map.put(sub, map.getOrDefault(sub, 0) + 1);
        if (map.get(sub) == 2) // ==保证统计所有且只记录一次
            res.add(root);
        return sub;
    }
}

C++

class Solution {
public:
    unordered_map<string, int> map;
    vector<TreeNode*> res;
    vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
        DFS(root);
        return res;
    }
    string DFS(TreeNode* root) {
        if (root == nullptr)
            return " ";
        string sub = "";
        sub += to_string(root->val); // 转换为字符串!!!
        sub += "-";
        sub += DFS(root->left);
        sub += DFS(root->right);
        if (map.count(sub))
            map[sub]++;
        else
            map[sub] = 1;
        if (map[sub] == 2) // ==保证统计所有且只记录一次
            res.emplace_back(root);
        return sub;
    }
};

Rust

use std::rc::Rc;
use std::cell::RefCell;
use std::collections::HashMap;
impl Solution {
    pub fn find_duplicate_subtrees(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
        let mut res = Vec::new();
        fn DFS(root: &Option<Rc<RefCell<TreeNode>>>, map: &mut HashMap<String, i32>, res: &mut Vec<Option<Rc<RefCell<TreeNode>>>>) -> String {
            if root.is_none() {
                return " ".to_string();
            }
            let sub = format!("{}-{}{}", root.as_ref().unwrap().borrow().val, DFS(&root.as_ref().unwrap().borrow().left, map, res), DFS(&root.as_ref().unwrap().borrow().right, map, res));
            *map.entry(sub.clone()).or_insert(0) += 1;
            if map[&sub] == 2 { // ==保证统计所有且只记录一次
                res.push(root.clone());
            }
            sub            
        }
        DFS(&root, &mut HashMap::new(), &mut res);
        res
    }
}

思路二:DFS+三元组

Java

class Solution {
    Map<String, Pair<Integer, Integer>> map = new HashMap<String, Pair<Integer, Integer>>();
    List<TreeNode> res = new ArrayList<>();
    int flag = 0;
    public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
        DFS(root);
        return res;
    }
    public int DFS(TreeNode root) {
        if (root == null)
            return 0;  
        int[] tri = {root.val, DFS(root.left), DFS(root.right)};
        String sub = Arrays.toString(tri); // 当前子树
        if (map.containsKey(sub)) { // 已统计过
            int key = map.get(sub).getKey();
            int cnt = map.get(sub).getValue();
            map.put(sub, new Pair<Integer, Integer>(key, ++cnt));
            if (cnt == 2) // ==保证统计所有且只记录一次
                res.add(root);
            return key;
        }
        else { // 首次出现
            map.put(sub, new Pair<Integer, Integer>(++flag, 1));
            return flag;
        }
    }
}

C++

class Solution {
public:
    unordered_map<string, pair<int, int>> map;
    vector<TreeNode*> res;
    int flag = 0;
    vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
        DFS(root);
        return res;
    }
    int DFS(TreeNode* root) {
        if (root == nullptr)
            return 0;
        string sub = to_string(root->val) + to_string(DFS(root->left)) + to_string(DFS(root->right)); // 当前子树
        if (auto cur = map.find(sub); cur != map.end()) { // 已统计过
            int key = cur->second.first;
            int cnt = cur->second.second;
            map[sub] = {key, ++cnt};
            if (cnt == 2) // ==保证统计所有且只记录一次
                res.emplace_back(root);
            return key;
        } 
        else { // 首次出现
            map[sub] = {++flag, 1};
            return flag;
        }
    }
};

Rust

use std::rc::Rc;
use std::cell::RefCell;
use std::collections::HashMap;
impl Solution {
    pub fn find_duplicate_subtrees(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
        let mut res = Vec::new();
        fn DFS(root: &Option<Rc<RefCell<TreeNode>>>, sub_flag: &mut HashMap<String, i32>, flag_cnt: &mut HashMap<i32, i32>, res: &mut Vec<Option<Rc<RefCell<TreeNode>>>>, flag: &mut i32) -> i32 {
            if root.is_none() {
                return 0;
            }
            let (lflag, rflag) = (DFS(&root.as_ref().unwrap().borrow().left, sub_flag, flag_cnt, res, flag), DFS(&root.as_ref().unwrap().borrow().right, sub_flag, flag_cnt, res, flag));
            let sub = format!("{}{}{}", root.as_ref().unwrap().borrow().val, lflag, rflag);
            if sub_flag.contains_key(&sub) { // 已统计过
                let key = sub_flag[&sub];
                let cnt = flag_cnt[&key] + 1;
                flag_cnt.insert(key, cnt);
                if cnt == 2 { // ==保证统计所有且只记录一次
                    res.push(root.clone());
                }
                key
            }
            else { // 首次出现
                *flag += 1;
                sub_flag.insert(sub, *flag);
                flag_cnt.insert(*flag, 1);
                *flag
            }
        }
        DFS(&root, &mut HashMap::new(), &mut HashMap::new(), &mut res, &mut 0);
        res
    }
}

总结

两种方法本质上都是基于哈希表,记录重复的子树结构并统计个数,在超过111时进行记录,不过思路二更巧妙地将冗长的字符串变为常数级的标识符。

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