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;}};