1. 题目
给定一个树，按中序遍历重新排列树，使树中最左边的结点现在是树的根，并且每个结点没有左子结点，只有一个右子结点。
示例 ：输入：[5,3,6,2,4,null,8,1,null,null,null,7,9]5/ \3 6/ \ \2 4 8/ / \
1 7 9输出：[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
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/increasing-order-search-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
2. 解题
- 二叉树中序遍历
class Solution {
public:TreeNode* increasingBST(TreeNode* root) {bool getroot = false;stack<TreeNode*> stk;TreeNode *newroot, *prev, *cur;while (root || !stk.empty()){while(root){stk.push(root);root = root->left;}root = stk.top();if(!getroot){newroot = new TreeNode(root->val);prev = newroot;getroot = true;}else{cur = new TreeNode(root->val);prev->right = cur;prev = cur;}stk.pop();root = root->right;}return newroot;}
};
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def increasingBST(self, root: TreeNode) -> TreeNode:prev = Nonehead = Nones = []while root or len(s)>0:while root:s.append(root)cur = rootroot = root.leftcur.left = Noneroot = s.pop()if prev:prev.right = rootelse:head = rootprev = rootroot = root.rightreturn head
36 ms 14.9 MB Python3
)
