[LeetCode] 255. Verify Preorder Sequence in Binary Search Tree 验证二叉搜索树的先序序列
Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree.
You may assume each number in the sequence is unique.
Follow up:
Could you do it using only constant space complexity?
给一个数组，验证是否为一个二叉搜索树的先序遍历出的序列。
二叉树的特点是：左<根<右，如果用中序遍历得到的结果就是有序数组，而先序遍历的结果就不是有序数组。
Python:
# Time: O(n)
# Space: O(h)
class Solution2:
# @param {integer[]} preorder
# @return {boolean}
def verifyPreorder(self, preorder):
low = float("-inf")
path = []
for p in preorder:
if p < low:
return False
while path and p > path[-1]:
low = path[-1]
path.pop()
path.append(p)
return True
Python:
# Time: O(n)
# Space: O(1)
class Solution:
# @param {integer[]} preorder
# @return {boolean}
def verifyPreorder(self, preorder):
low, i = float("-inf"), -1
for p in preorder:
if p < low:
return False
while i >= 0 and p > preorder[i]:
low = preorder[i]
i -= 1
i += 1
preorder[i] = p
return True
C++:
// Time: O(n)
// Space: O(h)
class Solution2 {
public:
bool verifyPreorder(vector<int>& preorder) {
int low = INT_MIN;
stack<int> path;
for (auto& p : preorder) {
if (p < low) {
return false;
}
while (!path.empty() && p > path.top()) {
// Traverse to its right subtree now.
// Use the popped values as a lower bound because
// we shouldn't come across a smaller number anymore.
low = path.top();
path.pop();
}
path.emplace(p);
}
return true;
}
};
C++:
// Time: O(n)
// Space: O(1)
class Solution {
public:
bool verifyPreorder(vector<int>& preorder) {
int low = INT_MIN, i = -1;
for (auto& p : preorder) {
if (p < low) {
return false;
}
while (i >= 0 && p > preorder[i]) {
low = preorder[i--];
}
preorder[++i] = p;
}
return true;
}
};
类似题目：
[LeetCode] 144. Binary Tree Preorder Traversal 二叉树的先序遍历
[LeetCode] 98. Validate Binary Search Tree 验证二叉搜索树
