leetcode binary tree upside down

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這題要收費了。只能網上看題目，我等屌絲也沒法OJ測試了。網上看了後發現其實並非獨創，其他的方也有類似的題。例如在這裡，CareerCup上

先用了遞迴的想法，

    TreeNode *ans;    TreeNode *helper156(TreeNode *root)    {        if (!root -> left && !root -> right)        {            ans = root; return root;        }        TreeNode *parent = UpsideDown(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;    }

 

在這裡提到如果是非遞迴的那麼和反轉鏈表差不多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;    }

不用遞迴還是挺妙的啊。

leetcode binary tree upside down