letcode之Recover Binary Search Tree-优快云博客

本文介绍了一种通过中序遍历找出并交换二叉搜索树中两个错误节点值的方法。首先进行中序遍历获取节点值，然后再次遍历找出需要交换的两个数值，并在树中修正这些值。

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这题首先是要找出来那2个数有问题。按照中序搜索排列之后，就可以找到哪2个有问题了。之后再遍历一次，将错误的数据恢复过来。代码如下：

# 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.
        """
        list1 = []
        def midorder(root):
            if not root:
                return root
            if root.left:
                midorder(root.left)
            list1.append(root.val)
            if root.right:
                midorder(root.right)
            return list1
        midorder(root)
        for i in range(len(list1) - 1):
            if list1[i] > list1[i + 1]:
                two1 = list1[i]
                break
        for i in range(1, len(list1)):
            if list1[i] < list1[i - 1]:
                two2 = list1[i]
       
        stack3 = [two1, two2]
        def change(x, num):
            if x.val in stack3:
                if x.val == stack3[0]:
                    x.val = stack3[1]
                else:
                    x.val = stack3[0]
                num = num + 1
                if num == 2:
                    return 
            for i in (x.left, x.right):
                if i:
                    change(i, num)
        change(root, 0)