# LC 94. Binary Tree Inorder Traversal ### [Problem link](https://leetcode.com/problems/binary-tree-inorder-traversal/) ###### tags: `leedcode` `easy` `python` `c++` `Binary Tree` Given the <code>root</code> of a binary tree, return the inorder traversal of its nodes' values. **Example 1:** <img alt="" src="https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg" style="width: 125px; height: 200px;" /> ``` Input: root = [1,null,2,3] Output: [1,3,2] ``` **Example 2:** ``` Input: root = [] Output: [] ``` **Example 3:** ``` Input: root = [1] Output: [1] ``` **Constraints:** - The number of nodes in the tree is in the range <code>[0, 100]</code>. - <code>-100 <= Node.val <= 100</code> **Follow up:** Recursive solution is trivial, could you do it iteratively? ## Solution 1 #### Python ##### recursive ```python= # 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]: res = [] def traversal(node: TreeNode): if not node: return traversal(node.left) res.append(node.val) traversal(node.right) traversal(root) return res ``` ##### iteration ```python= # 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]: res = [] stack = [] if root: stack.append(root) while stack: node = stack.pop() if node != None: if node.right: stack.append(node.right) stack.append(node) stack.append(None) if node.left: stack.append(node.left) else: node = stack.pop() res.append(node.val) return res ``` #### C++ ##### recursive ```cpp= /** * 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> res; void traversal(TreeNode *cur) { if (cur == nullptr) { return; } traversal(cur->left); res.push_back(cur->val); traversal(cur->right); } vector<int> inorderTraversal(TreeNode* root) { traversal(root); return res; } }; ``` ##### iteration ```cpp= /** * 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> res; stack<TreeNode*> st; if (root != nullptr) { st.push(root); } while (!st.empty()) { TreeNode *node = st.top(); st.pop(); if (node != nullptr) { if (node->right) { st.push(node->right); } st.push(node); st.push(nullptr); if (node->left) { st.push(node->left); } } else { node = st.top(); st.pop(); res.push_back(node->val); } } return res; } }; ``` >### Complexity >n = number of tree's nodes. >| | Time Complexity | Space Complexity | >| ----------- | --------------- | ---------------- | >| Solution 1 | O(n) | O(n) | ## Note [iteration reference](https://github.com/youngyangyang04/leetcode-master/blob/master/problems/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E7%BB%9F%E4%B8%80%E8%BF%AD%E4%BB%A3%E6%B3%95.md) [recursive reference](https://github.com/youngyangyang04/leetcode-master/blob/master/problems/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E8%BF%AD%E4%BB%A3%E9%81%8D%E5%8E%86.md)