{ "data": { "question": { "questionId": "2558", "questionFrontendId": "2471", "categoryTitle": "Algorithms", "boundTopicId": 1962149, "title": "Minimum Number of Operations to Sort a Binary Tree by Level", "titleSlug": "minimum-number-of-operations-to-sort-a-binary-tree-by-level", "content": "
You are given the root
of a binary tree with unique values.
In one operation, you can choose any two nodes at the same level and swap their values.
\n\nReturn the minimum number of operations needed to make the values at each level sorted in a strictly increasing order.
\n\nThe level of a node is the number of edges along the path between it and the root node.
\n\n\n
Example 1:
\n\n\nInput: root = [1,4,3,7,6,8,5,null,null,null,null,9,null,10]\nOutput: 3\nExplanation:\n- Swap 4 and 3. The 2nd level becomes [3,4].\n- Swap 7 and 5. The 3rd level becomes [5,6,8,7].\n- Swap 8 and 7. The 3rd level becomes [5,6,7,8].\nWe used 3 operations so return 3.\nIt can be proven that 3 is the minimum number of operations needed.\n\n\n
Example 2:
\n\n\nInput: root = [1,3,2,7,6,5,4]\nOutput: 3\nExplanation:\n- Swap 3 and 2. The 2nd level becomes [2,3].\n- Swap 7 and 4. The 3rd level becomes [4,6,5,7].\n- Swap 6 and 5. The 3rd level becomes [4,5,6,7].\nWe used 3 operations so return 3.\nIt can be proven that 3 is the minimum number of operations needed.\n\n\n
Example 3:
\n\n\nInput: root = [1,2,3,4,5,6]\nOutput: 0\nExplanation: Each level is already sorted in increasing order so return 0.\n\n\n
\n
Constraints:
\n\n[1, 105]
.1 <= Node.val <= 105
给你一个 值互不相同 的二叉树的根节点 root
。
在一步操作中,你可以选择 同一层 上任意两个节点,交换这两个节点的值。
\n\n返回每一层按 严格递增顺序 排序所需的最少操作数目。
\n\n节点的 层数 是该节点和根节点之间的路径的边数。
\n\n\n\n
示例 1 :
\n\n输入:root = [1,4,3,7,6,8,5,null,null,null,null,9,null,10]\n输出:3\n解释:\n- 交换 4 和 3 。第 2 层变为 [3,4] 。\n- 交换 7 和 5 。第 3 层变为 [5,6,8,7] 。\n- 交换 8 和 7 。第 3 层变为 [5,6,7,8] 。\n共计用了 3 步操作,所以返回 3 。\n可以证明 3 是需要的最少操作数目。\n\n\n
示例 2 :
\n\n输入:root = [1,3,2,7,6,5,4]\n输出:3\n解释:\n- 交换 3 和 2 。第 2 层变为 [2,3] 。 \n- 交换 7 和 4 。第 3 层变为 [4,6,5,7] 。 \n- 交换 6 和 5 。第 3 层变为 [4,5,6,7] 。\n共计用了 3 步操作,所以返回 3 。 \n可以证明 3 是需要的最少操作数目。\n\n\n
示例 3 :
\n\n输入:root = [1,2,3,4,5,6]\n输出:0\n解释:每一层已经按递增顺序排序,所以返回 0 。\n\n\n
\n\n
提示:
\n\n[1, 105]
。1 <= Node.val <= 105
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