Find Duplicate Subtrees问题及解法

问题描述：

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.

示例: 

        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.

问题分析：

我们考虑把树的每一个子树都序列化，序列化机构相同的子树即为我们要找的答案。这里我们采用map来存储子树相同序列化的结构及节点。

过程详见代码：

/** * 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*>> map;vector<TreeNode*> dups;serialize(root, map);for (auto it = map.begin(); it != map.end(); it++) {if (it->second.size() > 1) dups.push_back(it->second[0]);}return dups;}string serialize(TreeNode * node, unordered_map<string, vector<TreeNode*>> &map){if (!node) return "";string s = "(" + serialize(node->left, map) + to_string(node->val) + serialize(node->right, map) + ")";map[s].push_back(node);return s;}};