标签：modify instead ant ons ack void put python iter
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:
Approach #1: C++.[Recursive]
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void recoverTree(TreeNode* root) {
if (root == NULL) return;
helper(root);
int temp = first->val;
first->val = second->val;
second->val = temp;
}
private:
TreeNode* first = NULL;
TreeNode* second = NULL;
TreeNode* prev = NULL;
void helper(TreeNode* root) {
if (root == NULL) return;
helper(root->left);
if (prev != NULL && prev->val > root->val) {
if (first == NULL) first = prev;
second = root;
}
prev = root;
helper(root->right);
}
};
Approach #2: Java.[Iterator]
/**
* 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) {
if (root == null) return;
TreeNode first = null;
TreeNode second = null;
TreeNode prev = null;
Stack<TreeNode> stack = new Stack<>();
TreeNode curr = root;
while (!stack.isEmpty() || curr != null) {
if (curr != null) {
stack.push(curr);
curr = curr.left;
} else {
curr = stack.pop();
if (prev != null && prev.val > curr.val) {
if (first == null) first = prev;
second = curr;
}
prev = curr;
curr = curr.right;
}
}
int temp = first.val;
first.val = second.val;
second.val = temp;
}
}
Approach #3: Python. [Recursive]
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def recoverTree(self, root):
"""
:type root: TreeNode
:rtype: void Do not return anything, modify root in-place instead.
"""
self.first = None
self.second = None
self.prev = None
if root == None:
return;
self.helper(root);
temp = self.first.val
self.first.val = self.second.val
self.second.val = temp
def helper(self, root):
if root == None:
return;
self.helper(root.left)
if self.prev != None and self.prev.val > root.val:
if self.first == None:
self.first = self.prev
self.second = root
self.prev = root
self.helper(root.right)
99. Recover Binary Search Tree
标签：modify instead ant ons ack void put python iter
原文地址：https://www.cnblogs.com/ruruozhenhao/p/10092385.html
