CODE 39. Submission Details

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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?
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}".

public ArrayList<Integer> inorderTraversal(TreeNode root) {
	// Start typing your Java solution below
	// DO NOT write main() function
	if (null == root) {
		return new ArrayList<Integer>();
	}
	ArrayList<Integer> inorder = new ArrayList<Integer>();
	Stack<TreeNode> stack = new Stack<TreeNode>();
	TreeNode node = root;
	while (!stack.isEmpty() || null != node) {
		if (null == node) {
			node = stack.pop();
			inorder.add(node.val);
			node = node.right;
		} else {
			stack.push(node);
			node = node.left;
		}
	}
	return inorder;
}