94.Binary Tree Inorder Traversal

题目链接：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}".
 *
 */
public class BinaryTreeInorderTraversal {

	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}
	
//	67 / 67 test cases passed.
//	Status: Accepted
//	Runtime: 216 ms
//	Submitted: 1 minute ago
	
	//简单的中序遍历

	//递归版
    public List<Integer> inorderTraversal(TreeNode root) {
    	List<Integer> inorder= new ArrayList<Integer>();
    	inorderTraversal(root, inorder);
    	return inorder;
    }
    public void inorderTraversal(TreeNode root, List<Integer> inorder) {
    	if (root != null) {
    		inorderTraversal(root.left, inorder);
    		inorder.add(root.val);
    		inorderTraversal(root.right, inorder);
    	}
    }
    
    //遍历版
    public List<Integer> inorderTraversal1(TreeNode root) {
    	
    	List<Integer> inorder= new ArrayList<Integer>();
    	
    	Stack<TreeNode> stack = new Stack<TreeNode>();
    	
    	TreeNode p = root;
    	
    	while(!stack.isEmpty() || p != null) {
    		
    		if(p != null) {
    			stack.add(p);
    			p = p.left;
    		} else {
    			TreeNode node = stack.pop();
    			inorder.add(node.val);
    			p = node.right;
			}
    	}
    	return inorder;
    }
    
	public static void main(String[] args) {
		// TODO Auto-generated method stub

	}

}

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