LeetCode

我觉得这两道题没有技巧可言，唯有把逻辑弄清楚，再用代码把自己的逻辑写下来。

下面是这两道题的AC代码：

/**
    * Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
    * @param root
    */
    public void connect(TreeLinkNode root){
        if(root == null)
            return;
        root.next = null;
        if(root.left == null && root.right == null)
            return;
        TreeLinkNode next = root;
        while(next.left!=null && next.right!=null){
            TreeLinkNode temp = next.left;
            while(true)
            {
                if(next.left!=null&& next.right!=null)
                {
                    next.left.next = next.right;
                    next.right.next = next.next==null?null:next.next.left;
                }
                if(next.next == null)
                    break;
                next = next.next;
            }
            next = temp;
        }
    }
    /**
     * What if the given tree could be any binary tree? Would your previous solution still work?
     * @param root
     */
    public void connectAny(TreeLinkNode root){
        if(root == null)
            return ;
        //the root next pointer point to null;
        root.next = null;
        if(root.left == null && root.right == null)
            return ;
        //the current node to be handled
        TreeLinkNode curr = root;
        
        while(curr!=null){
            TreeLinkNode next =null;
            TreeLinkNode temp = null;
            while(true){
                //find the non-leaf node
                while(curr.left == null && curr.right == null){
                    if(curr.next == null)
                        break;
                    curr = curr.next;
                }
                if(next == null)
                    next = curr.left==null?(curr.right==null?null:curr.right):curr.left;
                //temp the node who need to populate the next pointer
                if(temp !=null)
                    temp.next = curr.left==null?(curr.right==null?null:curr.right):curr.left;
                //
                if(curr.left!=null && curr.right!=null)
                {
                    curr.left.next = curr.right;
                    temp = curr.right;
                }else if(curr.left!=null){
                    temp = curr.left;
                }else if(curr.right!=null){
                    temp = curr.right;
                }
                
                if(curr.next ==null){
                    if(temp!=null)
                        temp.next = null;
                    break;
                }
                curr = curr.next;
            }
            curr = next;
        }
    }