LeetCode：897. Increasing Order Search Tree

解题思路

Given a 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 1 right child.

Example 1:

Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

 5 / \  3 6 / \  \  2 4 8 / / \  1 7 9 Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9] 1 \  2 \  3 \  4 \  5 \  6 \  7 \  8 \  9

Note:

The number of nodes in the given tree will be between 1 and 100.
Each node will have a unique integer value from 0 to 1000.

解题思路 —— 分治求解

分治法。先分别对左/右子树做 Increasing 操作，然后将左右子树通过 root 节点连接起来。

AC 代码

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */
class Solution {
    TreeNode* increasingBST(TreeNode* root, TreeNode*& lastNode)
    {
        if(root == nullptr) return nullptr;
        TreeNode* leftTail = nullptr, *rightTail = nullptr;
        TreeNode* leftHead, *rightHead;

        // 对左子树做 increasingBST 操作
        leftHead = increasingBST(root->left, leftTail);
        if(leftHead != nullptr) leftTail->right = root; // 连接左子树
        else leftHead = root;

        // 对右子树做 increasingBST 操作
        rightHead = increasingBST(root->right, rightTail);
        root->right = rightHead; // 连接右子树
        root->left = nullptr;

        if(rightHead == nullptr) lastNode = root;
        else lastNode = rightTail;

        return leftHead;
    }
public:
    TreeNode* increasingBST(TreeNode* root) {
        TreeNode* lastNode;
        return increasingBST(root, lastNode);
    }
};