foolish fly fox's blog
--Stay hungry, stay foolish.
--Forever young, forever weeping.
https://leetcode.com/problems/recover-binary-search-tree/description/
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Example 1:
Input: [1,3,null,null,2]
1
/
3
\
2
Output: [3,1,null,null,2]
3
/
1
\
2
Example 2:
Input: [3,1,4,null,null,2]
3
/ \
1 4
/
2
Output: [2,1,4,null,null,3]
2
/ \
1 4
/
3
Follow up:
# 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 recoverTree(self, root): """ :type root: TreeNode :rtype: void Do not return anything, modify root in-place instead. """ if root==None: return record = [] cur_node, stack = root, [] while cur_node or len(stack): if cur_node: if cur_node.left: stack.append(cur_node) cur_node = cur_node.left else: record.append(cur_node) cur_node = cur_node.right else: record.append(stack.pop()) if record[-1].right: cur_node = record[-1].right wrong1 = wrong2 = None for i in range(1, len(record)): if wrong1==None and record[i-1].val>record[i].val: wrong1 = record[i-1] if wrong1 and record[i].val>wrong1.val: wrong2 = record[i-1] break else: wrong2 = record[-1] wrong1.val, wrong2.val = wrong2.val, wrong1.val