LeetCode(94)：二叉树的中序遍历

题干

给定一个二叉树，返回它的中序 遍历。

示例:

输入: [1,null,2,3]
1
\
2
/
3

输出: [1,3,2]

Code

• 这道题虽然一般，但重拾C++编程，收获到很多当时没注意的点，比如这种写法真的有醍醐灌顶之感
  TreeNode(int x) : val(x), left(NULL), right(NULL) {};
  想到当时不注意初始化问题而花费很多经历取解决Bug, 真的太生产力了

  typedef struct TreeNode {
  	int val;
  	TreeNode *left;
      TreeNode *right;
  	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  }TreeNode,*Tree;
   
  class Solution {
  public:
  	vector<int> inorderTraversal(TreeNode* root) {
  		vector<int> v;
  		helpInorderTraversal(root, v);
  		return v;
  	}
  	void helpInorderTraversal(TreeNode* root, vector<int> &v) {
  		if (root) {
  			helpInorderTraversal(root->left, v);
  			v.push_back(root->val);
  			helpInorderTraversal(root->right, v);
  		}
  	 }
  };
  

测试代码

#include<iostream>
#include<vector>

using namespace std;

// Definition for a binary tree node.
typedef struct TreeNode {
	int val;
	TreeNode *left;
    TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
}TreeNode,*Tree;
 
class Solution {
public:
	vector<int> inorderTraversal(TreeNode* root) {
		vector<int> v;
		helpInorderTraversal(root, v);
		return v;
	}
	void helpInorderTraversal(TreeNode* root, vector<int> &v) {
		if (root) {
			helpInorderTraversal(root->left, v);
			v.push_back(root->val);
			helpInorderTraversal(root->right, v);
		}
	 }
};

int main()
{
	Tree T = new TreeNode(1);
	Tree temp = T;
	temp->right = new TreeNode(2);
	temp = temp->right;
	temp->left = new TreeNode(3);

	Solution s;
	vector<int> v = s.inorderTraversal(T);
	for (auto i : v) {
		cout << i << " ";
	}
	
	return 0;
}