Title: Increasing Order Search Tree Source: leetcode.com
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 (Increasing order search tree).
Example 1:
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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.
Python solution
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''' https://leetcode.com/problems/increasing-order-search-tree/ ''' # 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 increasingBST(self, root): """ :type root: TreeNode :rtype: TreeNode """ if not root: return root #print root.val if not root.left and not root.right: return root left = None if root.left: left = self.increasingBST(root.left) right = self.increasingBST(root.right) if left: x = left while x.right: x = x.right x.right = root root.right = right root.left = None root = left else: root.right = right #print "returning ", root.val return root |