力扣labuladong——一刷day71

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前言

二叉树的递归分为「遍历」和「分解问题」两种思维模式，这道题需要用到「分解问题」的思维模式。

一、力扣109. 有序链表转换二叉搜索树

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * 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 {
    public TreeNode sortedListToBST(ListNode head) {
        List<Integer> list = new ArrayList<>();
        ListNode p = head;
        while(p != null){
            list.add(p.val);
            p = p.next;
        }
        return fun(list,0,list.size()-1);
    }
    public TreeNode fun(List<Integer> list, int low,int high){
        if(low > high){
            return null;
        }
        int mid = (low+high)/2;
        TreeNode cur = new TreeNode(list.get(mid));
        cur.left = fun(list,low,mid-1);
        cur.right = fun(list,mid+1,high);
        return cur;
    }
}

二、力扣1382. 将二叉搜索树变平衡

/**
 * 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 {
    List<Integer> list = new ArrayList<>();
    public TreeNode balanceBST(TreeNode root) {
        traverse(root);
        return fun(0,list.size()-1);
    }
    public TreeNode fun(int low, int high){
        if(low > high){
            return null;
        }
        int mid = (low+high)/2;
        TreeNode cur = new TreeNode(list.get(mid));
        cur.left = fun(low,mid-1);
        cur.right = fun(mid+1,high);
        return cur;
    }
    public void traverse(TreeNode root){
        if(root == null){
            return;
        }
        traverse(root.left);
        list.add(root.val);
        traverse(root.right);
    }
}