题目
1 class Solution { 2 public: 3 TreeNode* r1;TreeNode* r2; 4 bool isCousins(TreeNode* root, int x, int y) { 5 dfs(root,r1,x); 6 dfs(root,r2,y); 7 int h_x = Depth(root,x); 8 int h_y = Depth(root,y); 9 if(h_x == h_y && r1 != r2) return true; 10 return false; 11 } 12 void dfs(TreeNode* root,TreeNode* &r,int v){ 13 if(root!= NULL) { 14 dfs(root->left,r,v); 15 if(root->left &&root->left->val == v ) r = root; 16 if(root->right && root->right->val == v) r = root; 17 dfs(root->right,r,v); 18 } 19 } 20 int Depth(TreeNode* root,int v){ 21 if(root == NULL) return -1; 22 if(root->val == v) return 0; 23 //TreeNode* node; 24 queue<TreeNode*>q; 25 q.push(root); 26 int depth = 0; 27 while(!q.empty()){ 28 int num = q.size(); //一定要有这句，否则会影响下面循环次数 29 for(int i = 0;i < num;i++){ 30 TreeNode* node = q.front();q.pop(); 31 if(node->val == v) return depth; 32 if(node->left != NULL) q.push(node->left); 33 if(node->right != NULL) q.push(node->right); 34 } 35 depth++; 36 } 37 return depth; 38 } 39 };
注意：
1.依旧是函数的形参为指针情况，如果想要保存对指针的修改需要址传递 如dfs中 TreeNode* &r
2.第28行代码别忘记，一定要提前记录每一层的个数