[LeetCode] 99. 恢复二叉搜索树

题目链接 : https://leetcode-cn.com/problems/recover-binary-search-tree/

题目描述:

二叉搜索树中的两个节点被错误地交换。

请在不改变其结构的情况下，恢复这棵树。

示例:

示例 1:

输入: [1,3,null,null,2]

   1
  /
 3
  \
   2

输出: [3,1,null,null,2]

   3
  /
 1
  \
   2

示例 2:

输入: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

输出: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3

进阶:

使用 O(n) 空间复杂度的解法很容易实现。
你能想出一个只使用常数空间的解决方案吗？

思路:

这道题难点,是找到那两个交换节点,把它交换过来就行了.

这里我们二叉树搜索树的中序遍历(中序遍历遍历元素是递增的)

如下图所示, 中序遍历顺序是 4,2,3,1,我们只要找到节点4和节点1交换顺序即可!

这里我们有个规律发现这两个节点:

第一个节点,是第一个按照中序遍历时候前一个节点大于后一个节点,我们选取前一个节点,这里指节点4;

第二个节点,是在第一个节点找到之后, 后面出现前一个节点大于后一个节点,我们选择后一个节点,这里指节点1;

对于中序遍历,我们有两种方法.

方法一: 迭代

方法二: 递归

详细链接 关于树遍历.

代码:

方法一:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def recoverTree(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        firstNode = None
        secondNode = None
        pre = TreeNode(float("-inf"))

        stack = []
        p = root
        while p or stack:
            while p:
                stack.append(p)
                p = p.left
            p = stack.pop()
            
            if not firstNode and pre.val > p.val:
                    firstNode = pre
            if firstNode and pre.val > p.val:
                #print(firstNode.val,pre.val, p.val)
                secondNode = p
            pre = p
            p = p.right
        firstNode.val, secondNode.val = secondNode.val, firstNode.val

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public void recoverTree(TreeNode root) {
        Deque<TreeNode> stack = new LinkedList<>();
        TreeNode firstNode = null;
        TreeNode secondNode = null;
        TreeNode pre = new TreeNode(Integer.MIN_VALUE);
        TreeNode p = root;
        while (p != null || !stack.isEmpty()) {
            while (p != null) {
                stack.push(p);
                p = p.left;
            }
            p = stack.pop();
            if (firstNode == null && pre.val > p.val) firstNode = pre;
            if (firstNode != null && pre.val > p.val) secondNode = p;
            pre = p;
            p = p.right;
        }
        int tmp = firstNode.val;
        firstNode.val = secondNode.val;
        secondNode.val = tmp;
    }
}

思路二:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def recoverTree(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        self.firstNode = None
        self.secondNode = None
        self.preNode = TreeNode(float("-inf"))

        def in_order(root):
            if not root:
                return
            in_order(root.left)
            if self.firstNode == None and self.preNode.val >= root.val:
                self.firstNode = self.preNode
            if self.firstNode and self.preNode.val >= root.val:
                self.secondNode = root
            self.preNode = root
            in_order(root.right)

        in_order(root)
        self.firstNode.val, self.secondNode.val = self.secondNode.val, self.firstNode.val

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    TreeNode firstNode = null;
    TreeNode secondNode = null;
    TreeNode preNode = new TreeNode(Integer.MIN_VALUE);

    public void recoverTree(TreeNode root) {

        in_order(root);
        int tmp = firstNode.val;
        firstNode.val = secondNode.val;
        secondNode.val = tmp;
    }

    private void in_order(TreeNode root) {
        if (root == null) return;
        in_order(root.left);
        if (firstNode == null && preNode.val > root.val) firstNode = preNode;
        if (firstNode != null && preNode.val > root.val) secondNode = root;
        preNode = root;
        in_order(root.right);
    }
}

标签：TreeNode,val,self,二叉,99,firstNode,null,root,LeetCode
来源： https://www.cnblogs.com/powercai/p/11080707.html