LeetCode 94. Binary Tree Inorder Traversal--二叉树中序遍历--递归，迭代--C++,Python解法

题目地址：Binary Tree Inorder Traversal - LeetCode

Given a binary tree, return the inorder traversal of its nodes’ values.

Example:

Input: [1,null,2,3]1\2/3Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

经典的二叉树中序遍历，有迭代和递归2种做法，递归的比较简单。
Python解法如下：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def inorderTraversal(self, root: TreeNode) -> List[int]:def helper(root, l):if root is None:returnhelper(root.left,l)l.append(root.val)helper(root.right,l)l = []if root is None:return lhelper(root, l)return l

C++解法如下：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {public:vector<int> l;void helper(TreeNode *root) {if (root == nullptr) {return;}helper(root->left);l.push_back(root->val);helper(root->right);}vector<int> inorderTraversal(TreeNode *root) {l.clear();helper(root);return l;}
};

递归的解法总是比迭代简单，迭代解法如下：

Python解法如下：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def inorderTraversal(self, root: TreeNode) -> List[int]:if root == None:return []l = []res = []curr = rootwhile curr != None or l != []:while curr is not None:l.append(curr)curr = curr.leftcurr = l.pop(-1)res.append(curr.val)curr = curr.rightreturn res

时间复杂度为O(n)，空间复杂度为O(1)。

C++解法如下：

/*** Definition for a binary tree node.* 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) {vector<int> result;stack<TreeNode *> l;if (root == nullptr) {return result;}TreeNode *curr = root;while (curr != nullptr || !l.empty()) {while (curr != nullptr) {l.push(curr);curr = curr->left;}curr = l.top();l.pop();result.push_back(curr->val);curr = curr->right;}return result;}
};

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