## 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].

Analysis:
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.

Code(C++):


/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> res;
inOrder(root,res);
return res;
}
};


### Code(python):

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
# @param root, a tree node
# @return a list of integers
res = []
def iot(self, root):
if root == None:
return
else:
self.iot(root.left)
self.res.append(root.val)
self.iot(root.right)

def inorderTraversal(self, root):
self.res = []   #reset the result list
self.iot(root)
return self.res



#### 1 comment:

1. Excuse for my little Ads.
A solution by iteration rather then recursion and a solution by iteration with constant space can be found here