leetcode[156] binary tree upside down

这题要收费了。只能网上看题目，我等屌丝也没法OJ测试了。网上看了后发现其实并非独创，其他的方也有类似的题。例如在这里，CareerCup上

先用了递归的想法，

    TreeNode *ans;

    TreeNode *helper156(TreeNode *root)

    {

        if (!root -> left && !root -> right)

        {

            ans = root; return root;

        }

        TreeNode *parent = helper156(root -> left);

        parent -> left = root -> right;

        parent -> right = root;

        root -> left = NULL;

        root -> right = NULL;

        return parent -> right;

    }

    TreeNode *UpsideDown(TreeNode *root)

    {

        if (root)

            helper156(root);

        return ans;

    }

 

在这里提到如果是非递归的那么和反转链表差不多Reverse Linked List II

    TreeNode *UpsideDown2(TreeNode *root)

    {

        if (!root -> left && !root -> right) return root;

        TreeNode *parent = NULL, *parentRight = NULL, *p = root;

        while(p)

        {

            TreeNode *left = p -> left;

            p -> left = parentRight;

            parentRight = p -> right;

            p -> right = parent;

            parent = p;

            p = left;

        }

        return parent;

    }

不用递归还是挺妙的啊。