思路：
实现：
1 class Solution 2 { 3 public: 4 void dfs(TreeNode* root, vector<int>& v) 5 { 6 if (root == NULL) return; 7 dfs(root->left, v); 8 v.push_back(root->val); 9 dfs(root->right, v); 10 } 11 TreeNode* canMerge(vector<TreeNode*>& trees) 12 { 13 map<int, TreeNode*> mp; 14 map<TreeNode*, TreeNode*> par; 15 int cnt = 0; 16 for (auto it: trees) 17 { 18 if (it->left) 19 { 20 int tmp = it->left->val; 21 if (mp.count(tmp)) return NULL; 22 mp[it->left->val] = it->left; 23 par[it->left] = it; 24 cnt++; 25 } 26 if (it->right) 27 { 28 int tmp = it->right->val; 29 if (mp.count(tmp)) return NULL; 30 mp[it->right->val] = it->right; 31 par[it->right] = it; 32 cnt++; 33 } 34 } 35 TreeNode* root = NULL; 36 for (auto it: trees) 37 { 38 int tmp = it->val; 39 if (!mp.count(tmp)) 40 { 41 if (root) return NULL; 42 else root = it; 43 } 44 else 45 { 46 TreeNode* x = mp[tmp]; 47 TreeNode* p = par[x]; 48 if (x == p->left) p->left = it; 49 else p->right = it; 50 delete(x); 51 } 52 } 53 vector<int> res; 54 dfs(root, res); 55 int n = res.size(); 56 if (n != cnt + 1) return NULL; 57 bool flg = true; 58 for (int i = 0; i < n - 1; i++) 59 { 60 if (res[i] >= res[i + 1]) { flg = false; break; } 61 } 62 if (!flg) return NULL; 63 return root; 64 } 65 };