LeetCode OJ

这题的提示中给出，linked list是原来二叉树的先序遍历。所以我用一个栈暂时存储每个节点的右子树（如果它有右子树的话），然后把左子树交换到右子树，左节点置为null，之后迭代下去，当遇到叶子节点时，从栈中pop出一个子树，再继续，直到遇到叶子节点且栈中为空。

下面是AC代码：

 /**
     * Given a binary tree, flatten it to a linked list in-place.
     * If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
     * @param root
     */
    public void flatten(TreeNode root){
        if(root == null || root.left==null && root.right == null)
            return;
        //the stack is for the right children
        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
        TreeNode temp = root;
        while(true){
            if(temp.left!=null)
            {
                if(temp.right!=null)
                    stack.push(temp.right);
                temp.right = temp.left;
                temp.left = null;
                
            }
            if(temp.right != null)
                temp = temp.right;
            if(temp.left == null && temp.right == null){
                if(stack.isEmpty())
                    break;
                temp.right = stack.pop();
                temp = temp.right;
            }
        }
        
    }