我实现了 Morris 遍历来解决 LeetCode 问题 98。验证二叉搜索树:
给定二叉树的
,确定它是否是有效的二叉搜索树(BST)。root
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool isValidBST(TreeNode* root) {
long aa = LONG_MIN;
long bb = LONG_MIN;
TreeNode* cur = root;
while(cur){
if(cur->left == NULL){
aa = max(aa, bb);
bb = cur->val;
if(aa != LONG_MIN && aa >= bb) return false;
cur = cur->right;
}
else{
TreeNode* prev = cur->left;
while(prev->right && prev->right != cur){
prev = prev->right;
}
if(prev->right == NULL){
prev->right = cur;
cur = cur->left;
}
else{
prev->right = NULL;
aa = max(aa, bb);
bb = cur->val;
if(aa != LONG_MIN && aa >= bb) return false;
cur = cur->right;
}
}
}
return true;
}
};
这是一个标准的 [Morris 遍历实现],我刚刚添加了一个检查(在代码中的两个位置),将节点的值与其中序前驱的值进行比较,以查看该树是否是有效的 BST:
if(aa != LONG_MIN && aa >= bb) return false;
在 LeetCode 提交此代码,会产生以下错误:
AddressSanitizer:DEADLYSIGNAL
=================================================================
==22==ERROR: AddressSanitizer: stack-overflow on address 0x7ffcecdecff8 (pc 0x5627c311bad9 bp 0x7ffcecded010 sp 0x7ffcecded000 T0)
#0 0x5627c311bad9 in __TreeNodeUtils__::freeTreeHelper(TreeNode*) (solution+0x18cad9)
#1 0x5627c311bb00 in __TreeNodeUtils__::freeTreeHelper(TreeNode*) (solution+0x18cb00)
#2 0x5627c311badd in __TreeNodeUtils__::freeTreeHelper(TreeNode*) (solution+0x18cadd)
#3 0x5627c311bb00 in __TreeNodeUtils__::freeTreeHelper(TreeNode*) (solution+0x18cb00)
[...]
#243 0x5627c311bb00 in __TreeNodeUtils__::freeTreeHelper(TreeNode*) (solution+0x18cb00)
#244 0x5627c311badd in __TreeNodeUtils__::freeTreeHelper(TreeNode*) (solution+0x18cadd)
#245 0x5627c311bb00 in __TreeNodeUtils__::freeTreeHelper(TreeNode*) (solution+0x18cb00)
#246 0x5627c311badd in __TreeNodeUtils__::freeTreeHelper(TreeNode*) (solution+0x18cadd)
SUMMARY: AddressSanitizer: stack-overflow (solution+0x18cad9) in __TreeNodeUtils__::freeTreeHelper(TreeNode*)
==22==ABORTING
如果我删除检查 BST 属性的行:
if(aa != LONG_MIN && aa >= bb) return false;
...然后就不再有错误了。为什么这一行会出现上述错误?
该错误不会发生在您的函数中,而是发生在 LeetCode 平台代码中,该代码在函数返回后释放分配给树的内存。
莫里斯遍历会改变树,暂时在其中引入循环。这意味着,如果您在中途的某个地方中止该遍历,则可能会在树中留下这些循环,这是函数的调用者不希望发生的情况。当该调用代码尝试清理时(请注意堆栈跟踪中提到的
freeTreeHelper
),它会沿着这些循环运行,大概使用递归函数,该函数永远不会停止遍历其中一个循环上的节点。这最终会导致堆栈溢出。
为了避免这种情况发生,请确保在返回调用者之前将树恢复到其原始状态(或至少没有循环)。
一种方法是继续 Morris 遍历到底,并使用布尔变量跟踪无效的 BST 状态。
像这样:
class Solution {
public:
bool isValidBST(TreeNode* root) {
long aa = LONG_MIN;
long bb = LONG_MIN;
bool valid = true; // Add this boolean to keep track of BST validity.
TreeNode* cur = root;
while(cur){
if(cur->left == NULL){
aa = max(aa, bb);
bb = cur->val;
// Don't exit, but just keep track of the validity
valid &= aa == LONG_MIN || aa < bb;
cur = cur->right;
}
else{
TreeNode* prev = cur->left;
while(prev->right && prev->right != cur){
prev = prev->right;
}
if(prev->right == NULL){
prev->right = cur;
cur = cur->left;
}
else{
prev->right = NULL;
aa = max(aa, bb);
bb = cur->val;
// Don't exit, but just keep track of the validity
valid &= aa == LONG_MIN || aa < bb;
cur = cur->right;
}
}
}
return valid; // return the boolean we have kept updated
}
};