标签：struct when solution start while tty ken des python
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Example 1:
Input: [1,3,null,null,2] 1 / 3 2 Output: [3,1,null,null,2] 3 / 1 2
Example 2:
Input: [3,1,4,null,null,2] 3 / 1 4 / 2 Output: [2,1,4,null,null,3] 2 / 1 4 / 3
Follow up:
二叉搜索树的2个元素被错误的交换了，在不改变结构的情况下恢复二叉搜索树。follow up: 用常量空间。
Java:
public class Solution {
TreeNode firstElement = null;
TreeNode secondElement = null;
// The reason for this initialization is to avoid null pointer exception in the first comparison when prevElement has not been initialized
TreeNode prevElement = new TreeNode(Integer.MIN_VALUE);
public void recoverTree(TreeNode root) {
// In order traversal to find the two elements
traverse(root);
// Swap the values of the two nodes
int temp = firstElement.val;
firstElement.val = secondElement.val;
secondElement.val = temp;
}
private void traverse(TreeNode root) {
if (root == null)
return;
traverse(root.left);
// Start of "do some business",
// If first element has not been found, assign it to prevElement (refer to 6 in the example above)
if (firstElement == null && prevElement.val >= root.val) {
firstElement = prevElement;
}
// If first element is found, assign the second element to the root (refer to 2 in the example above)
if (firstElement != null && prevElement.val >= root.val) {
secondElement = root;
}
prevElement = root;
// End of "do some business"
traverse(root.right);
}
Python:
# Time: O(n)
# Space: O(1)
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
def __repr__(self):
if self:
serial = []
queue = [self]
while queue:
cur = queue[0]
if cur:
serial.append(cur.val)
queue.append(cur.left)
queue.append(cur.right)
else:
serial.append("#")
queue = queue[1:]
while serial[-1] == "#":
serial.pop()
return repr(serial)
else:
return None
Python:
class Solution(object):
# @param root, a tree node
# @return a tree node
def recoverTree(self, root):
return self.MorrisTraversal(root)
def MorrisTraversal(self, root):
if root is None:
return
broken = [None, None]
pre, cur = None, root
while cur:
if cur.left is None:
self.detectBroken(broken, pre, cur)
pre = cur
cur = cur.right
else:
node = cur.left
while node.right and node.right != cur:
node = node.right
if node.right is None:
node.right =cur
cur = cur.left
else:
self.detectBroken(broken, pre, cur)
node.right = None
pre = cur
cur = cur.right
broken[0].val, broken[1].val = broken[1].val, broken[0].val
return root
def detectBroken(self, broken, pre, cur):
if pre and pre.val > cur.val:
if broken[0] is None:
broken[0] = pre
broken[1] = cur
C++:
// O(n) space complexity
class Solution {
public:
void recoverTree(TreeNode *root) {
vector<TreeNode*> list;
vector<int> vals;
inorder(root, list, vals);
sort(vals.begin(), vals.end());
for (int i = 0; i < list.size(); ++i) {
list[i]->val = vals[i];
}
}
void inorder(TreeNode *root, vector<TreeNode*> &list, vector<int> &vals) {
if (!root) return;
inorder(root->left, list, vals);
list.push_back(root);
vals.push_back(root->val);
inorder(root->right, list, vals);
}
};
C++:
// Still O(n) space complexity
class Solution {
public:
TreeNode *pre;
TreeNode *first;
TreeNode *second;
void recoverTree(TreeNode *root) {
pre = NULL;
first = NULL;
second = NULL;
inorder(root);
if (first && second) swap(first->val, second->val);
}
void inorder(TreeNode *root) {
if (!root) return;
inorder(root->left);
if (!pre) pre = root;
else {
if (pre->val > root->val) {
if (!first) first = pre;
second = root;
}
pre = root;
}
inorder(root->right);
}
};
C++:
// O(1) space complexity
class Solution {
public:
void recoverTree(TreeNode *root) {
TreeNode *first = NULL, *second = NULL, *parent = NULL;
TreeNode *cur, *pre;
cur = root;
while (cur) {
if (!cur->left) {
if (parent && parent->val > cur->val) {
if (!first) first = parent;
second = cur;
}
parent = cur;
cur = cur->right;
} else {
pre = cur->left;
while (pre->right && pre->right != cur) pre = pre->right;
if (!pre->right) {
pre->right = cur;
cur = cur->left;
} else {
pre->right = NULL;
if (parent->val > cur->val) {
if (!first) first = parent;
second = cur;
}
parent = cur;
cur = cur->right;
}
}
}
if (first && second) swap(first->val, second->val);
}
};
[LeetCode] 99. Recover Binary Search Tree 复原二叉搜索树
标签：struct when solution start while tty ken des python
原文地址：https://www.cnblogs.com/lightwindy/p/9795766.html
