2013.12.31 18:26
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1 \ 2 / 3
return [1,3,2]
.
Note: 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.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / \ 2 3 / 4 \ 5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}"
.
Solution:
Here is the definition for in-order traversal of a binary tree, click here.
Time comlexity is O(n), space complexity is O(1), where n is the number of nodes in the tree.
Accepted code:
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> inorderTraversal(TreeNode *root) { 13 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 res.clear(); 16 if(root == nullptr){ 17 return res; 18 } 19 20 inorder(root); 21 22 return res; 23 } 24 private: 25 vector<int> res; 26 void inorder(TreeNode *root) { 27 if(root == nullptr){ 28 return; 29 }else{ 30 if(root->left != nullptr){ 31 inorder(root->left); 32 } 33 res.push_back(root->val); 34 if(root->right != nullptr){ 35 inorder(root->right); 36 } 37 } 38 } 39 };