99. 恢复二叉搜索树

原题

https://leetcode.cn/problems/recover-binary-search-tree/description/

思路

中序遍历 + 双指针

复杂度

时间：O(n)
空间：O(n)

Python代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        l = []

        def dfs(root):
            if not root:
                return 
            dfs(root.left)
            l.append(root)
            dfs(root.right)

        dfs(root)
        x = None 
        y = None 
        pre = l[0]
        for i in range(1, len(l)):
            if pre.val > l[i].val:
                y = l[i]
                if not x:
                    x = pre
            pre = l[i]
        if x and y:
            x.val, y.val = y.val, x.val 
     

Go代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func recoverTree(root *TreeNode) {
	var l []*TreeNode
	// 匿名函数
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Left)
		l = append(l, root)
		dfs(root.Right)
	}

	dfs(root)
	var x, y *TreeNode
	pre := l[0]
	for i := 1; i < len(l); i++ {
		if pre.Val > l[i].Val {
			y = l[i]
			if x == nil {
				x = pre
			}
		}
		pre = l[i]
	}
	if x != nil && y != nil {
		x.Val, y.Val = y.Val, x.Val
	}
}