Binary Tree Inorder Traversal @ leetcode

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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?

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}".

» Solve this problem

利用栈来存储结点。注意while的判断。

Further Thoughts:
The above solutions require the help of a stack to do in-order traversal. Is it possible to do in-order traversal without a stack?

The answer is yes, it’s possible. There’s 2 possible ways that I know of:

1. By adding a parent pointer to the data structure, this allows us to return to a node’s parent (Credits to my friend who provided this solution to me). To determine when to print a node’s value, we would have to determine when it’s returned from. If it’s returned from its left child, then you would print its value then traverse to its right child, on the other hand if it’s returned from its right child, you would traverse up one level to its parent.
2. By using a Threaded Binary Tree. Read the article: Threaded Binary Tree on Wikipedia for more information.

Depth-first

• Pre-order traversal sequence: F, B, A, D, C, E, G, I, H (root, left, right) 
• In-order traversal sequence: A, B, C, D, E, F, G, H, I (left, root, right) 
• Post-order traversal sequence: A, C, E, D, B, H, I, G, F (left, right, root) 

Breadth-first

• Level-order traversal sequence: F, B, G, A, D, I, C, E, H