Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
Example
Tags Expand
解题思路：
思路一：平衡树调整。第一步建立一个单向树，然后使用平衡树的调整。逻辑比较绕，虽然完成了代码，但是大数据时，测试超时。
思路二：保存所有节点到数组（List集合），然后使用二分法，依次建立所有的头节点和左子树，右子树。
思路三：参考网上解法，核心还是使用二分法，但是比较巧妙的一点是在确定所有元素个数和划分左右子树后，能够实现指针的一次遍历，读取全部数据，无需多次将数据读出保存。
思路一：
/**
* Definition for ListNode.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int val) {
* this.val = val;
* this.next = null;
* }
* }
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param head: The first node of linked list.
* @return: a tree node
*/
public TreeNode sortedListToBST(ListNode head) {
if (head == null)
return null;
TreeNode root = null;
while (head != null) {
int v = head.val;
head = head.next;
TreeNode node = new TreeNode(v);
node.left = root;
root = node;
}
return adjustNode(root);
}
TreeNode adjustNode(TreeNode node) {
if (node != null) {
int l = getLevel(node.left);
int r = getLevel(node.right);
while (l - r >= 2) {
TreeNode tmp = node.left;
if (tmp.right == null) {
node.left = null;
tmp.right = node;
node = tmp;
} else {
TreeNode lmax = tmp.right;
while (lmax.right != null) {
tmp = lmax;
lmax = lmax.right;
}
tmp.right = null;
lmax.left = node.left;
lmax.right = node;
node.left = null;
node = lmax;
}
l = getLevel(node.left);
r = getLevel(node.right);
}
while (r - l >= 2) {
TreeNode tmp = node.right;
if (tmp.left == null) {
node.right = null;
tmp.left = node;
node = tmp;
} else {
TreeNode lmax = tmp.left;
while (lmax.left != null) {
tmp = lmax;
lmax = lmax.left;
}
tmp.left = null;
lmax.right = node.right;
lmax.left = node;
node.right = null;
node = lmax;
}
l = getLevel(node.left);
r = getLevel(node.right);
}
node.left = adjustNode(node.left);
node.right = adjustNode(node.right);
}
return node;
}
int getLevel(TreeNode node) {
if (node == null)
return 0;
else
return Math.max(getLevel(node.left) + 1, getLevel(node.right) + 1);
}
}
解法二：
/**
* Definition for ListNode.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int val) {
* this.val = val;
* this.next = null;
* }
* }
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param head: The first node of linked list.
* @return: a tree node
*/
public TreeNode sortedListToBST(ListNode head) {
if (head == null)
return null;
ArrayList<Integer> list = new ArrayList<Integer>();
while (head != null) {
list.add(head.val);
head = head.next;
}
TreeNode root = getNode(list, 0, list.size() - 1);
return root;
}
private TreeNode getNode(ArrayList<Integer> list, int x1, int x2) {
if (x1 == x2)
return new TreeNode(list.get(x1));
else if (x1 > x2)
return null;
else {
int mid = (x1 + x2) / 2;
TreeNode node = new TreeNode(list.get(mid));
node.left = getNode(list, x1, mid - 1);
node.right = getNode(list, mid + 1, x2);
return node;
}
}
}
思路三：（网上解法）
public TreeNode sortedListToBST(ListNode head) {
if(head == null)
return null;
ListNode cur = head;
int count = 0;
while(cur!=null)
{
cur = cur.next;
count++;
}
ArrayList<ListNode> list = new ArrayList<ListNode>();
list.add(head);
return helper(list,0,count-1);
}
private TreeNode helper(ArrayList<ListNode> list, int l, int r)
{
if(l>r)
return null;
int m = (l+r)/2;
TreeNode left = helper(list,l,m-1);
TreeNode root = new TreeNode(list.get(0).val);
root.left = left;
list.set(0,list.get(0).next);
root.right = helper(list,m+1,r);
return root;
}