leetcode 652. Find Duplicate Subtrees Map记录查询 + 深度优先遍历DFS

约定不等于承诺〃 2022-06-03 07:09 141阅读 0赞

Given a binary tree, return all duplicate subtrees. For each kind of duplicate subtrees, you only need to return the root node of any one of them.

Two trees are duplicate if they have the same structure with same node values.

Example 1:  
1  
/ \\  
2 3  
/ / \\  
4 2 4  
/  
4  
The following are two duplicate subtrees:  
2  
/  
4  
and  
4  
Therefore, you need to return above trees’ root in the form of a list.

本题最初的想法就是一个结点一个节点的查询，不过这个需要很多的计算，会很费时间，后来发现一个直接递归的做法，**用到了前序遍历（后序遍历也可以，中序遍历不可以），还有数组序列化，并且建立序列化跟其出现次数的映射，这样如果我们得到某个结点的序列化字符串，而该字符串正好出现的次数为1，说明之前已经有一个重复树了**，我们将当前结点存入结果res，这样保证了多个重复树只会存入一个结点，参见代码如下：

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    /*
    struct TreeNode 
    {
         int val;
         TreeNode *left;
         TreeNode *right;
         TreeNode(int x) : val(x), left(NULL), right(NULL) {}
    };
    */
    
    class Solution 
    {
    public:
        vector<TreeNode*> res;
        map<string, int> mmp;
        vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) 
        {
            dfs(root);
            return res;
        }
    
        string dfs(TreeNode* root)
        {
            if (root == NULL)
                return "#";
            else
            {
                string left = dfs(root->left);
                string right = dfs(root->right);
                string id = to_string(root->val) + left + right;
                if (mmp.find(id) == mmp.end())
                    mmp[id] = 1;
                else
                {
                    if (mmp[id] == 1)
                        res.push_back(root);
                    mmp[id] += 1;
                }
                return id;
            }
        }
    };