Problem: 865. Smallest Subtree with all the Deepest Nodes 具有所有最深节点的最小子树
解题过程
耗时100%，记录所有最长的路径，也就是到叶节点的路径，然后遍历所有最长路径，统计每个节点出现的次数，和最大值，第一条路径内最后等于最大值的节点就是所求。
Code
/** * 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 { public: vector<vector<TreeNode*>> tr; vector<TreeNode*> tmp; int mx = INT_MIN; void dfs(TreeNode* root) { if(root==nullptr) return; tmp.push_back(root); if(root->left == nullptr && root->right == nullptr) { if((int)tmp.size() > mx) { tr.clear(); tr.push_back(tmp); mx = (int)tmp.size(); } else if((int)tmp.size() == mx) { tr.push_back(tmp); } tmp.pop_back(); return; } dfs(root->left); dfs(root->right); tmp.pop_back(); } TreeNode* subtreeWithAllDeepest(TreeNode* root) { dfs(root); if(tr.size() == 0) return nullptr; unordered_map<TreeNode*, int> ump; int mxmx = INT_MIN; for(int i = 0; i < tr.size(); i++) { for(int j = 0; j < tr[i].size(); j++) { ump[tr[i][j]]++; mxmx = max( mxmx, ump[tr[i][j]] ); } } TreeNode* pre = nullptr; for(int i = 0; i < tr[0].size(); i++) { if(ump[tr[0][i]]!=mxmx) { break; } pre = tr[0][i]; } return pre; } };
