Leetcode 897. Increasing Order Search Tree (二叉树遍历和分治好题)

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已于 2024-02-07 15:23:26 修改 · 591 阅读

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于 2024-02-06 15:40:21 首次发布

Increasing Order Search Tree
Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.

Example 1:

Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
Example 2:

Input: root = [5,1,7]
Output: [1,null,5,null,7]

Constraints:

The number of nodes in the given tree will be in the range [1, 100].
0 <= Node.val <= 1000

解法1：遍历。节点新生成。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
   
   
public:
    TreeNode* increasingBST(TreeNode* root) {
   
   
        helper(root);
        return head;
    }
private:
    TreeNode *head = NULL, *node = NULL;