[Leetcode] Same Tree
1最容易想到的递归调用
1 bool isSameTree(TreeNode *r1, TreeNode *r2) 2 { 3 if(r1 == NULL && r2 == NULL) return true; //终止条件 4 if(r1 == NULL || r2 == NULL) return false;//减枝，我觉得也是终止条件 5 6 if(r1->val == r2->val 7 && isSameTree(r1->left, r2->left) 8 && isSameTree(r1->right, r2->right)) //递归，三方合并 9 return true; 10 return false; 11 }
2 迭代，受下面前序方法的启发
前序：
1 class Solution { 2 public: 3 vector<int> preorderTraversal(TreeNode *root) { 4 vector<int> result; 5 const TreeNode *p; 6 stack<const TreeNode *> s; 7 p = root; 8 if (p != nullptr) s.push(p); 9 while (!s.empty()) { 10 p = s.top(); 11 s.pop(); 12 result.push_back(p->val); 13 if (p->right != nullptr) s.push(p->right); 14 if (p->left != nullptr) s.push(p->left); 15 } 16 return result; 17 } 18 };
迭代：
1 class Solution { 2 public: 3 bool isSameTree(TreeNode *r1, TreeNode* r2) { 4 5 if(r1 == NULL && r2 == NULL) return true; 6 if(r1 == NULL || r2 == NULL) return false; 7 8 stack<TreeNode*> stack; 9 stack.push(r1); 10 stack.push(r2); 11 12 TreeNode *p1; 13 TreeNode *p2; 14 15 while(!stack.empty()) 16 { 17 p1 = stack.top(); 18 stack.pop(); 19 p2 = stack.top(); 20 stack.pop(); 21 22 if(p1 == NULL && p2 == NULL) continue;//都为空时无法调用p1->right，p1->left 23 if(p1 == NULL || p2 == NULL) return false; 24 if(p1->val != p2->val) return false; 25 26 stack.push(p1->right); 27 stack.push(p2->right); 28 29 stack.push(p1->left); 30 stack.push(p2->left); 31 32 } 33 34 return true; 35 } 36 37 };
