[LeetCode] Binary Tree Upside Down 二叉樹的上下顛倒
Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.
For example:
Given a binary tree {1,2,3,4,5},
1
/ \
2 3
/ \
4 5
return the root of the binary tree [4,5,2,#,#,3,1].
4
/ \
5 2
/ \
3 1
這道題讓我們把一棵二叉樹上下顛倒一下,而且限制了右節點要麼為空要麼一定會有對應的左節點。上下顛倒後原來二叉樹的最左子節點變成了根節點,其對應的右節點變成了其左子節點,其父節點變成了其右子節點,相當於順時針旋轉了一下。對於一般樹的題都會有迭代和遞迴兩種解法,這道題也不例外,那麼我們先來看看遞迴的解法。對於一個根節點來說,我們的目標是將其左子節點變為根節點,右子節點變為左子節點,原根節點變為右子節點,那麼我們首先判斷這個根節點是否存在,且其有沒有左子節點,如果不滿足這兩個條件的話,直接返回即可,不需要翻轉操作。那麼我們不停的對左子節點呼叫遞迴函式,直到到達最左子節點開始翻轉,翻轉好最左子節點後,開始回到上一個左子節點繼續翻轉即可,直至翻轉完整棵樹,參見程式碼如下:
解法一:
// Recursion class Solution { public: TreeNode *upsideDownBinaryTree(TreeNode *root) { if (!root || !root->left) return root; TreeNode *l = root->left, *r = root->right; TreeNode *res = upsideDownBinaryTree(l); l->left = r; l->right = root; root->left = NULL; root->right = NULL; return res; } };
下面我們來看迭代的方法,和遞迴方法相反的時,這個是從上往下開始翻轉,直至翻轉到最左子節點,參見程式碼如下:
解法二:
// Iterative class Solution { public: TreeNode *upsideDownBinaryTree(TreeNode *root) { TreeNode *cur = root, *pre = NULL, *next = NULL, *tmp = NULL; while (cur) { next = cur->left; cur->left = tmp; tmp = cur->right; cur->right = pre; pre = cur; cur = next; } return pre; } };
參考資料: