LeetCode algorithm binary tree—108. Convert ordered array to binary search tree

Table of contents

108. Convert ordered array to binary search tree

Code:

operation result: 

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You are given an array of integers  nums in which the elements are sorted in  ascending  order. Please convert it into a  height-balanced  binary search tree.

A height-balanced  binary tree is a binary tree that satisfies "the absolute value of the height difference between the left and right subtrees of each node does not exceed 1".

Example 1:

Input: nums = [-10,-3,0,5,9]
 Output: [0,-3,9,-10,null,5]
 Explanation: [0,-10,5,null,-3,null ,9] will also be considered the correct answer:

Example 2:

Input: nums = [1,3]
 Output: [3,1]
 Explanation: [1,null,3] and [3,1] are both highly balanced binary search trees.

hint:

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums Arrange  in  strict increasing order

Code:

/**
 * 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 {
    // 「每个节点的左右两个子树的高度差的绝对值不超过 1 」的二叉树
    // 即将数组看作中序遍历[左根右]，那么数组中间就看做根节点，
    // 递归处理：根据中序遍历确定根节点，然后递归确定左子树，右子树
    public TreeNode sortedArrayToBST(int[] nums) {
        // 给出数组，本次需确定树的数组范围，初始为0——数组结束
        return fun(nums,0,nums.length-1);
    }
    TreeNode fun(int[] nums,int beg,int end){
        // 递归终止条件：该部分数组内容处理完了
        if(beg>end) return null;
        // 按照中间（靠右）为根节点
        int mid=(end-beg+1)/2+beg;
        // 确定当前范围根节点
        TreeNode root=new TreeNode(nums[mid]);
        // 递归
        root.left=fun(nums,beg,mid-1);
        root.right=fun(nums,mid+1,end);
        // 返回当前树
        return root;
    }
}

operation result: