LeetCode #94 Binary Tree Inorder Traversal

問題概要

入力値：root (num array)
出力値：num array
return the inorder traversal
問題のリンク

入力例

root: [1,null,2,3]
answer: [1,3,2]

解答例1

計算量：O(n)
DFS using recursion
Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        def search(root, result):
            if root:
                search(root.left, result)
                result.append(root.val)
                search(root.right, result)
        result = []
        search(root, result)
        return result

Runtime: 14ms
Beats: 55.25%

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        search(root, result);
        return result;
    }

    void search(TreeNode* root, vector<int>& result) {
        if (root != nullptr) {
            search(root->left, result);
            result.push_back(root->val);
            search(root->right, result);
        }
    }
};

Runtime: 0ms
Beats: 100%