897. Increasing Order Search Tree
1. Question:
897. Increasing Order Search Tree
url: https://leetcode.com/problems/increasing-order-search-tree/description/
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.
2. Solution:
# Definition for a binary tree node. class TreeNode(object): def __init__(self, x): self.val = x self.left = None self.right = None class Solution(object): def inOrder(self, root, path): if root is None: return self.inOrder(root.left, path) path.append(root) self.inOrder(root.right, path) def increasingBST(self, root): """ :type root: TreeNode :rtype: TreeNode """ path_node = [] self.inOrder(root, path_node) if len(path_node) <= 0: return None re_root = path_node[0] tail_node = re_root for node in path_node[1:]: tail_node.left = None tail_node.right = node tail_node = node tail_node.left = None tail_node.right = None return re_root
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