[Leetcode] 652. Find Duplicate Subtrees 解题报告

题目：

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

思路：

发现利用序列化是个非常好的思路：子树相同的充要条件是其序列化结果也相同。因此我们定义一个从string到vector<TreeNode*>的哈希表，然后对每个非空节点进行序列化，并将结果加入哈希表中。如果某个string对应的数组大小大于1，就将其数组中的一个元素加入结果集中。这样算法的时间复杂度只有O(n)。

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
        unordered_map<string, vector<TreeNode*>> hash;
        serialize(root, hash);
        vector<TreeNode*> ret;
        for (auto it = hash.begin(); it != hash.end(); ++it) {
            if (it->second.size() > 1) {
                ret.push_back(it->second[0]);
            }
        }
        return ret;
    }
private:
    string serialize(TreeNode *node, unordered_map<string, vector<TreeNode*>> &hash) {
        if (!node) {
            return "";
        }
        string s = "(" + serialize(node->left, hash) 
                       + to_string(node->val) 
                       + serialize(node->right, hash) + ")";
        hash[s].push_back(node);
        return s;
    }
};