7.4 - hard - 23

99. Recover Binary Search Tree
这道题因为需要inplace来做， 所以很明显的又是针对recursion traversal的一种变形

# 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.
        """
        # 很显然要用inorder
        self.prev = None
        self.first = None
        self.second = None

        self.inorder(root)
        self.first.val, self.second.val = self.second.val, self.first.val
            
    
    def inorder(self, root):
        if not root:
            return
        
        self.inorder(root.left)
        if not self.prev or root.val > self.prev.val:
            self.prev = root
        elif not self.first:
            self.first = self.prev
            self.second = root
        else:
            self.second = root
        self.inorder(root.right)