LintCode: Binary Tree Inorder Traversal
C++,递归,辅助函数
1 /** 2 * Definition of TreeNode: 3 * class TreeNode { 4 * public: 5 * int val; 6 * TreeNode *left, *right; 7 * TreeNode(int val) { 8 * this->val = val; 9 * this->left = this->right = NULL; 10 * } 11 * } 12 */ 13 class Solution { 14 /** 15 * @param root: The root of binary tree. 16 * @return: Inorder in vector which contains node values. 17 */ 18 public: 19 vector<int> inorderTraversal(TreeNode *root) { 20 // write your code here 21 vector<int> result; 22 if (root == NULL) { 23 return result; 24 } else { 25 inorderCore(root, result); 26 } 27 return result; 28 } 29 void inorderCore(TreeNode *root, vector<int> &result) { 30 if (root->left != NULL) { 31 inorderCore(root->left, result); 32 } 33 result.push_back(root->val); 34 if (root->right != NULL) { 35 inorderCore(root->right, result); 36 } 37 } 38 };
C++,递归
1 /** 2 * Definition of TreeNode: 3 * class TreeNode { 4 * public: 5 * int val; 6 * TreeNode *left, *right; 7 * TreeNode(int val) { 8 * this->val = val; 9 * this->left = this->right = NULL; 10 * } 11 * } 12 */ 13 class Solution { 14 /** 15 * @param root: The root of binary tree. 16 * @return: Inorder in vector which contains node values. 17 */ 18 public: 19 vector<int> inorderTraversal(TreeNode *root) { 20 // write your code here 21 vector<int> result; 22 if (root == NULL) { 23 return result; 24 } 25 if (root->left != NULL) { 26 vector<int> left = inorderTraversal(root->left); 27 result.reserve(result.size() + left.size()); 28 result.insert(result.end(), left.begin(), left.end()); 29 } 30 result.push_back(root->val); 31 if (root->right != NULL) { 32 vector<int> right = inorderTraversal(root->right); 33 result.reserve(result.size() + right.size()); 34 result.insert(result.end(), right.begin(), right.end()); 35 } 36 return result; 37 } 38 };
C++,非递归
1 /** 2 * Definition of TreeNode: 3 * class TreeNode { 4 * public: 5 * int val; 6 * TreeNode *left, *right; 7 * TreeNode(int val) { 8 * this->val = val; 9 * this->left = this->right = NULL; 10 * } 11 * } 12 */ 13 class Solution { 14 /** 15 * @param root: The root of binary tree. 16 * @return: Inorder in vector which contains node values. 17 */ 18 public: 19 vector<int> inorderTraversal(TreeNode *root) { 20 // write your code here 21 vector<int> result; 22 if (root == NULL) { 23 return result; 24 } 25 stack<TreeNode *> sta; 26 TreeNode *cur = root; 27 while (cur != NULL || !sta.empty()) { 28 while(cur != NULL) { 29 sta.push(cur); 30 cur = cur->left; 31 } 32 cur = sta.top(); 33 sta.pop(); 34 result.insert(result.end(), cur->val); 35 cur = cur->right; 36 } 37 return result; 38 } 39 };
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作者:
ZH奶酪(张贺)
邮箱:
cheesezh@qq.com
出处:
http://www.cnblogs.com/CheeseZH/
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