Recover Binary Search Tree
題意:
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
解題思路:
對於bst作中序遍歷,使用兩個list,分別紀錄節點與節點的值,最後再根據對應的位置 assign 對應的值,時間複雜度為O(NlogN),空間複雜度為O(N),其程式碼如下:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
List<TreeNode> nodes = new ArrayList<TreeNode>();
List<Integer> values = new ArrayList<Integer>();
public void recoverTree(TreeNode root) {
if (root == null) {
return;
}
inorderTraversal(root);
Collections.sort(values);
for (int i = 0; i < nodes.size(); i++) {
nodes.get(i).val = values.get(i);
}
}
public void inorderTraversal(TreeNode root) {
if (root == null) {
return;
}
inorderTraversal(root.left);
nodes.add(root);
values.add(root.val);
inorderTraversal(root.right);
}
}
另有O(1)空間複雜度與O(N)時間複雜度的解法