题目
给定一个二叉树的根节点 root ，返回它的 中序 遍历。
示例 1：
输入：root = [1,null,2,3] 输出：[1,3,2]
示例 2：
输入：root = [] 输出：[]
示例 3：
输入：root = [1] 输出：[1]
示例 4：
输入：root = [1,2] 输出：[2,1]
示例 5：
输入：root = [1,null,2] 输出：[1,2]
解题:
方法一：递归
python解法
# 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: TreeNode) -> List[int]: def inorder(root): if not root: return inorder(root.left) res.append(root.val) inorder(root.right) res = [] inorder(root) return res
可以找到规律，将先序遍历改成中序遍历只需要将res.append(root.val)放到中间即可
c++解法
class Solution { public: void inorder(TreeNode*root,vector<int>& res){ if(!root) return; inorder(root->left,res); res.push_back(root->val); inorder(root->right,res); } vector<int> inorderTraversal(TreeNode* root) { vector<int> res; inorder(root,res); return res; } };
方法二：迭代
python解法
# 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: TreeNode) -> List[int]: if not root: return [] cur,stack,res = root,[],[] while stack or cur: while cur: stack.append(cur) cur = cur.left cur = stack.pop() res.append(cur.val) cur = cur.right return res
c++解法
class Solution { public: vector<int> inorderTraversal(TreeNode* root) { if(!root) return {}; vector<int> res; stack<TreeNode*> stack; TreeNode* cur=root; while(!stack.empty()||cur){ while(cur){ stack.push(cur); cur=cur->left; } cur=stack.top(); stack.pop(); res.push_back(cur->val); cur=cur->right; } return res; } };
