LeetCode - Binary Tree Inorder Traversal

Binary Tree Inorder Traversal

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:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        res.clear();
        if(root == nullptr){
            return res;
        }
        
        inorder(root);
        
        return res;
    }
private:
    vector<int> res;
    void inorder(TreeNode *root) {
        if(root == nullptr){
            return;
        }else{
            if(root->left != nullptr){
                inorder(root->left);
            }
            res.push_back(root->val);
            if(root->right != nullptr){
                inorder(root->right);
            }
        }
    }
};