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99.recover-binary-search-tree.cpp
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98 lines (91 loc) · 2.64 KB
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// Tag: Tree, Depth-First Search, Binary Search Tree, Binary Tree
// Time: O(N)
// Space: O(H)
// Ref: -
// Note: -
// You are given the root of a binary search tree (BST), where the values of exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.
//
// Example 1:
//
//
// Input: root = [1,3,null,null,2]
// Output: [3,1,null,null,2]
// Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.
//
// Example 2:
//
//
// Input: root = [3,1,4,null,null,2]
// Output: [2,1,4,null,null,3]
// Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.
//
//
// Constraints:
//
// The number of nodes in the tree is in the range [2, 1000].
// -231 <= Node.val <= 231 - 1
//
//
// Follow up: A solution using O(n) space is pretty straight-forward. Could you devise a constant O(1) space solution?
/**
* 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:
void recoverTree(TreeNode* root) {
TreeNode* pre = nullptr;
TreeNode* first = nullptr;
TreeNode* second = nullptr;
inorder(root, pre, first, second);
swap(first->val, second->val);
}
void inorder(TreeNode* node, TreeNode* &pre, TreeNode* &first, TreeNode* &second) {
if (!node) {
return;
}
inorder(node->left, pre, first, second);
if (pre && pre->val > node->val) {
if (!first) {
first = pre;
}
second = node;
}
pre = node;
inorder(node->right, pre, first, second);
}
};
class Solution {
public:
void recoverTree(TreeNode* root) {
TreeNode* pre = nullptr;
TreeNode* first = nullptr;
TreeNode* second = nullptr;
stack<TreeNode *> st;
TreeNode* cur = root;
while (cur || !st.empty()) {
while (cur) {
st.push(cur);
cur = cur->left;
}
cur = st.top();
st.pop();
if (pre && pre->val > cur->val) {
if (!first) {
first = pre;
}
second = cur;
}
pre = cur;
cur = cur->right;
}
swap(first->val, second->val);
}
};