94 - Binary Tree Inorder Traversal

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

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as  "{1,2,3,#,#,4,#,#,5}".

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思路分析：

二叉树的中序遍历。

Classic tree operation, recursion is a straightforward idea to solve this problem.
Recursively do:
(1) Visit left child
(2) Output current node
(3) Visit right child
if current node is empty, return.

/*
*/

#include "stdafx.h"
#include <iostream>
#include <vector>

using namespace std;

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

class Solution_094_BinaryTreeInorderTraversal
{
public:
	void inOrder(TreeNode *root, vector<int> &res)
	{
		if (root != NULL)
		{
			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;
	}
};