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Binary Tree Inorder Traversal
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?
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 5The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}".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 14 vector<int> result;15 inOrder(root, result);16 return result;17 18 }19 20 void inOrder(TreeNode *root, vector<int> &result)21 {22 if(root != NULL)23 {24 inOrder(root->left, result);25 result.push_back(root->val);26 inOrder(root->right, result);27 }28 }29 };
Binary Tree Inorder Traversal
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