力扣每日一题2022-09-05中等题：寻找重复的子树

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题目描述

寻找重复的子树

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思路

DFS+哈希表

将每一棵子树都序列化称一个字符串，并保证相同的子树会被序列化成相同的子串、不同的子树会被序列化成不同的子串。只要使用一个哈希表存储所有子树的序列化结果，如果某个结果出现超过一次，就能发现一类重复子树。
序列化二叉树的方法可以使用层次遍历，也可以使用递归进行序列化。这里使用深度优先搜索的方法递归序列化二叉树并存入哈希表，遍历时遇到重复的子树序列化结果就可以加入结果数组中。

Python实现

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findDuplicateSubtrees(self, root: Optional[TreeNode]) -> List[Optional[TreeNode]]:
        def dfs(node):
            if not node:
                return ""
            serial = "".join([str(node.val), "(", dfs(node.left), ")(", dfs(node.right), ")"])
            if (tree := seen.get(serial, None)):
                repeat.add(tree)
            else:
                seen[serial] = node
            return serial

        seen = dict()
        repeat = set()
        dfs(root)
        return list(repeat)

Java实现

/**
 * 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;
 *     }
 * }
 */
class Solution {

    Map<String, TreeNode> seen = new HashMap<>();
    Set<TreeNode> repeat = new HashSet<>();

    public String dfs(TreeNode node) {
        if (node == null) {
            return "";
        }
        StringBuilder sb = new StringBuilder();
        sb.append(node.val);
        sb.append("(");
        sb.append(dfs(node.left));
        sb.append(")(");
        sb.append(dfs(node.right));
        sb.append(")");
        String serial = sb.toString();
        if (seen.containsKey(serial)) {
            repeat.add(seen.get(serial));
        } else {
            seen.put(serial, node);
        }
        return serial;
    }

    public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
        dfs(root);
        return new ArrayList<TreeNode>(repeat);
    }
}

C++实现

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    unordered_map<string, TreeNode*> seen;
    unordered_set<TreeNode*> repeat;

public:
    string dfs(TreeNode* node) {
        if (!node) {
            return "";
        }
        string serial = to_string(node->val) + "(" + dfs(node->left) + ")(" + dfs(node->right) + ")";
        if (auto tree = seen.find(serial); tree != seen.end()) {
            repeat.insert(tree->second);
        } else {
            seen[serial] = node;
        }
        return serial;
    }

    vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
        dfs(root);
        return {repeat.begin(), repeat.end()};
    }
};