problem
solution
另外建立兩棵樹，分別代表原先的樹
- 節點數
- 總和
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57class Solution {
public:
TreeNode * Sumtree(TreeNode *root){
if(!root) return nullptr;
// postorder
TreeNode *ret = new TreeNode(0);
if(root->left){
ret->left = Sumtree(root->left);
ret->val += ret->left->val;
}
if(root->right) {
ret->right = Sumtree(root->right);
ret->val += ret->right->val;
}
ret->val += root->val;
return ret;
}
TreeNode * CountNode(TreeNode *root){
if(!root) return nullptr;
TreeNode *ret = new TreeNode(1);
if(root->left){
ret->left = CountNode(root->left);
ret->val += ret->left->val;
}
if(root->right){
ret->right = CountNode(root->right);
ret->val += ret->right->val;
};
return ret;
}
void postorder(TreeNode* root){
if(!root) return;
postorder(root->left);
postorder(root->right);
cout<<root->val<<" ";
}
void averageOfSubtree(int &ret , TreeNode* root, TreeNode *sum, TreeNode * count){
if(!root) return ;
if(sum->val/ count->val == root->val) ret++;
averageOfSubtree(ret, root->left, sum->left, count->left);
averageOfSubtree(ret, root->right, sum->right, count->right);
}
int averageOfSubtree(TreeNode* root) {
// Node(node->val), number of node, sum
TreeNode *sum = Sumtree(root);
// postorder(sum);
TreeNode *count = CountNode(root);
// cout<<endl;
// postorder(count);
int ret = 0;
averageOfSubtree(ret,root, sum, count);
return ret;
}
};
簡化
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28/**
* 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:
int cnt = 0;
pair<int,int> dfs(TreeNode *root){
if(!root) return {0,0};
auto left = dfs(root->left), right = dfs(root->right);
int sum = left.first + right.first + root->val;
int count = 1 + left.second + right.second;
if(root->val == sum / count) cnt++;
return {sum, count};
}
int averageOfSubtree(TreeNode* root) {
auto p = dfs(root);
return cnt;
}
};
analysis
- time complexity
O(n) - space complexity
O(1)