leetcode 94. Binary Tree Inorder Traversal 二叉树中序递归遍历

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

For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].

这个题很简单，就是二叉树的中序遍历。

代码如下：

import java.util.ArrayList;
import java.util.List;

/*class TreeNode 
{
    int val;
     TreeNode left;
     TreeNode right;
     TreeNode(int x) { val = x; }
}*/

/*
 * 二叉树的只中序遍历
 * */
public class Solution
{
    List res=new ArrayList();
    public List inorderTraversal(TreeNode root) 
    {
        if(root==null) 
            return res;
        ByReci(root);
        return res;
    }
    private void ByReci(TreeNode root)
    {
        if(root ==null )
            return;
        else
        {
            ByReci(root.left);
            res.add(root.val);
            ByReci(root.right);
        }

    }
}

就是二叉树的中序遍历，直接DFS深度优先遍历即可

代码如下：

#include
#include 

using namespace std;

/*
struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
*/

class Solution 
{
public:
    vector<int> res;
    vector<int> inorderTraversal(TreeNode* root) 
    {
        getAll(root);
        return res;
    }

    void getAll(TreeNode* root)
    {
        if (root == NULL)
            return;
        else
        {
            getAll(root->left);
            res.push_back(root->val);
            getAll(root->right);
        }
    }
};