Binary Tree Preorder Traversal
2014.1.14 02:36
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1 \ 2 / 3
return [1,2,3]
.
Note: Recursive solution is trivial, could you do it iteratively?
Solution1:
The recursive version of preorder traversal, everybody knows what to do.
Time and space complexities are both O(n), where n is the number of nodes in the tree.
Accepted code:
1 // 1WA, 1RE, 1AC 2 /** 3 * Definition for binary tree 4 * struct TreeNode { 5 * int val; 6 * TreeNode *left; 7 * TreeNode *right; 8 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 9 * }; 10 */ 11 class Solution { 12 public: 13 vector<int> preorderTraversal(TreeNode *root) { 14 // IMPORTANT: Please reset any member data you declared, as 15 // the same Solution instance will be reused for each test case. 16 result.clear(); 17 // 1WA here, this sentence missing 18 preorder(root); 19 return result; 20 } 21 private: 22 vector<int> result; 23 24 void preorder(TreeNode *root) { 25 // 1RE here, if(root) 26 if(root == nullptr){ 27 return; 28 } 29 result.push_back(root->val); 30 preorder(root->left); 31 preorder(root->right); 32 } 33 };
Solution2:
Iterative version, not so easy as the last one.
Time and space complexities are still O(n).
Accepted code:
1 class Solution { 2 public: 3 vector<int> preorderTraversal(TreeNode *root) { 4 result.clear(); 5 6 if(root == nullptr){ 7 return result; 8 } 9 10 vstack.clear(); 11 cstack.clear(); 12 vstack.push_back(root); 13 cstack.push_back(0); 14 result.push_back(root->val); 15 while(vstack.size() > 0){ 16 if(cstack[cstack.size() - 1] == 0){ 17 ++cstack[cstack.size() - 1]; 18 if(vstack[vstack.size() - 1]->left != nullptr){ 19 cstack.push_back(0); 20 vstack.push_back(vstack[vstack.size() - 1]->left); 21 result.push_back(vstack[vstack.size() - 1]->val); 22 } 23 }else if(cstack[cstack.size() - 1] == 1){ 24 ++cstack[cstack.size() - 1]; 25 if(vstack[vstack.size() - 1]->right != nullptr){ 26 cstack[cstack.size() - 1] = 0; 27 vstack[vstack.size() - 1] = vstack[vstack.size() - 1]->right; 28 result.push_back(vstack[vstack.size() - 1]->val); 29 } 30 }else{ 31 cstack.pop_back(); 32 vstack.pop_back(); 33 } 34 } 35 36 return result; 37 } 38 private: 39 vector<int> result; 40 vector<TreeNode *> vstack; 41 vector<int> cstack; 42 };
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