94. Binary Tree Inorder Traversal
給定一個二叉樹的根節點，返回其節點值的中序遍歷。
範例 1：
輸入：root = [1, null, 2, 3] 輸出：[1, 3, 2]
Python
from typing import Optional, List
# Definition for a binary tree node.
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
result = []
self.inorderHelper(root, result)
return result
def inorderHelper(self, node: Optional[TreeNode], result: List[int]):
if node:
self.inorderHelper(node.left, result) # Traverse left subtree
result.append(node.val) # Visit node
self.inorderHelper(node.right, result) # Traverse right subtree
16.15MB, 44ms
C++
#include <vector>
using namespace std;
// Use the pre-defined TreeNode struct
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> result;
inorderHelper(root, result);
return result;
}
private:
void inorderHelper(TreeNode* node, vector<int>& result) {
if (node) {
inorderHelper(node->left, result); // Traverse left subtree
result.push_back(node->val); // Visit node
inorderHelper(node->right, result); // Traverse right subtree
}
}
};
9.8MB, 0ms
Javascript
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
/**
* @param {TreeNode} root
* @return {number[]}
*/
var inorderTraversal = function(root) {
const result = [];
const inorderHelper = (node) => {
if (node !== null) {
inorderHelper(node.left); // Traverse left subtree
result.push(node.val); // Visit node
inorderHelper(node.right); // Traverse right subtree
}
};
inorderHelper(root);
return result;
};
49.24MB, 42ms
