# 94. Binary Tree Inorder Traversal

Given a binary tree, return theinordertraversal of its nodes' values.

For example:
Given binary tree`[1,null,2,3]`,

``````   1
\
2
/
3
``````

return`[1,3,2]`.

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

Trivial Solution: According to the definition of Inorder (O(n) time and O(n) space, for the function call stack)

``````class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {

inorderTraversalHelper(root);
}

void inorderTraversalHelper(TreeNode * root){
if(!root) return;
inorderTraversalHelper(root->left);
inorderTraversalHelper(root->right);
}
};
``````

Thoughts

1. Using Stack to keep track path
2. *Morris traversal

Code Using Stack: (O(n) time and O(n) space)

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
stack<TreeNode *> path;
public:
vector<int> inorderTraversal(TreeNode* root) {

traverse(root);
}

void traverse(TreeNode* cur){
while(cur || !path.empty()){
// first to the left
while(cur){
path.push(cur);
cur = cur -> left;
}
// left done , do cur
cur = path.top(); path.pop();
// exploring right
cur = cur->right;
}
}
};
``````

Code with Morris Traversal: O(n) time and O(1) space (two assisting pointers)!

psudo-code (Thanks monkeykingyan's solution)

``````Step1. Initialize current as root
Step2. While current is not NULL
If current does not have left child
b. Go to the right, i.e., current = current.right
Else
a. In current's left subtree, make current the right child of the rightmost node
b. Go to this left child, i.e., current = current.left
``````

Code

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
TreeNode *cur = root;
while(cur){

if(cur->left){
TreeNode* pre = cur -> left;
while(pre -> right) pre = pre -> right;
pre->right = cur;
TreeNode * temp = cur;
cur = cur -> left;
temp -> left = NULL;
}
else{
cur = cur -> right;
}
}