099J. Recover Binary Search Tree
这个题还真不好写。 思想就是,一个排好序的数组里,如果你只发现了一个 peak, 那么说明 这个 peak 和 下一位的位置被调换了。 如果你发现了两个peak, 说明 第一个peak 和 第二个 peak 的 next 被调换了。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
ArrayList<TreeNode> inorder;
public void recoverTree(TreeNode root) {
this.inorder = new ArrayList<>();
getInorder(root);
TreeNode prev = null;
TreeNode next = null;
for (int i = 0; i < inorder.size() - 1; i++) {
if (inorder.get(i).val > inorder.get(i + 1).val) {
if (prev == null) {
prev = inorder.get(i);
next = inorder.get(i + 1);
} else {
next = inorder.get(i + 1);
}
}
}
int temp = prev.val;
prev.val = next.val;
next.val = temp;
}
private void getInorder(TreeNode node) {
if (node == null) return;
getInorder(node.left);
inorder.add(node);
getInorder(node.right);
}
}
运用上述的思想,也可以轻松的写出递归的解法。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private TreeNode prev;
private TreeNode next;
private TreeNode parent;
public void recoverTree(TreeNode root) {
prev = null;
next = null;
parent = null;
helper(root);
int temp = prev.val;
prev.val = next.val;
next.val = temp;
}
private void helper(TreeNode node) {
if (node == null) return;
helper(node.left);
if (parent != null) {
if (parent.val > node.val) {
if (prev == null) {
prev = parent;
next = node;
} else {
next = node;
}
}
}
parent = node;
helper(node.right);
}
}
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