[LeetCode] Binary Tree Inorder Traversal Solution
2013-01-17 15:15
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Given a binary tree, return the inorder[/i] traversal of its nodes' values.For example:
Given binary tree
[Thoughts]
For recursion version, it's very easy to write.
But for iterative version, we need a stack to help.
Iteration version
Given binary tree
{1,#,2,3},
1 \ 2 / 3return
[1,3,2].Note:[/b] Recursive solution is trivial, could you do it iteratively?confused what
"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.» Solve this problem
[Thoughts]
For recursion version, it's very easy to write.
But for iterative version, we need a stack to help.
Recursion version [code]1: vector<int> inorderTraversal(TreeNode *root) { 2: // Start typing your C/C++ solution below 3: // DO NOT write int main() function 4: vector<int> result; 5: inorderTra(root, result); 6: return result; 7: } 8: void inorderTra(TreeNode* node, vector<int> &result) 9: { 10: if(node == NULL) 11: { 12: return; 13: } 14: inorderTra(node->left, result); 15: result.push_back(node->val); 16: inorderTra(node->right, result); 17: }
Iteration version
1: vector<int> inorderTraversal(TreeNode *root) { 2: // Start typing your C/C++ solution below 3: // DO NOT write int main() function 4: vector<TreeNode*> sta; 5: vector<int> result; 6: if(root == NULL) return result; 7: TreeNode* node =root; 8: while(sta.size()>0 || node!=NULL) 9: { 10: while(node!=NULL) 11: { 12: sta.push_back(node); 13: node = node->left; 14: } 15: node= sta.back(); 16: sta.pop_back(); 17: result.push_back(node->val); 18: node =node->right; 19: } 20: return result; 21: }
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