LeetCode: Convert Sorted List to Binary Search Tree 解题报告
Convert Sorted List to Binary Search Tree
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
SOLUTION 1:
public TreeNode sortedListToBST1(ListNode head) {
ListNode fast = head;
ListNode slow = head; ListNode pre = head; if (head == null) {
return null;
} TreeNode root = null;
if (head.next == null) {
root = new TreeNode(head.val);
root.left = null;
root.right = null;
return root;
} // get the middle node.
while (fast != null && fast.next != null) {
fast = fast.next.next; // record the node before the SLOW.
pre = slow;
slow = slow.next;
} // cut the list to two parts.
pre.next = null;
TreeNode left = sortedListToBST1(head);
TreeNode right = sortedListToBST1(slow.next); root = new TreeNode(slow.val);
root.left = left;
root.right = right; return root;
}
SOLUTION 2:
public TreeNode sortedListToBST(ListNode head) {
if (head == null) {
return null;
} int size = ;
ListNode cur = head;
while (cur != null) {
size++;
cur = cur.next;
} CurrNode curNode = new CurrNode(head);
return sortedListToBSTHelp(curNode, size);
} public class CurrNode {
ListNode node; CurrNode(ListNode node) {
this.node = node;
}
} // when the recursion is done, the curr node should point to the node
// which is the next of the block.
public TreeNode sortedListToBSTHelp(CurrNode curr, int size) {
if (size <= ) {
return null;
} TreeNode left = sortedListToBSTHelp(curr, size/); // because we want to deal with the right block.
TreeNode root = new TreeNode(curr.node.val);
curr.node = curr.node.next; TreeNode right = sortedListToBSTHelp(curr, size - - size/); root.left = left;
root.right = right; return root;
}
SOLUTION 3:
并且这个dfs有一个作用 会把指针移动到这个要建的树的下一个位置
经过这一行 cur就移动到了中间
我们建立 一个根 TreeNode root = new TreeNode(curNode.val);
把cur移动到下一个 curNode = curNode.next;
再用递归建立右树
root.right = right;
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; next = null; }
* }
*/
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
ListNode curNode = null; public TreeNode sortedListToBST(ListNode head) {
if (head == null) {
return null;
} int size = ;
ListNode cur = head;
while (cur != null) {
size++;
cur = cur.next;
} curNode = head;
return dfs(head, size);
} // Use the size to control.
public TreeNode dfs(ListNode head, int size) {
if (size <= ) {
return null;
} TreeNode left = dfs(head, size / );
TreeNode root = new TreeNode(curNode.val); // move the current node to the next place.
curNode = curNode.next;
TreeNode right = dfs(curNode, size - size / - ); root.left = left;
root.right = right; return root;
}
}
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