【LeetCode题目记录-9】排序后的数组生成平衡的二叉搜索树

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Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

【分析1-原创】中间值作为根节点，左边的中间值作为左孩子，右边的中间值作为右孩子。一直递归探底即可。

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode sortedArrayToBST(int[] num) {        if(num==null) return null;        //获取中间的值作为根节点        TreeNode root=getMiddle(num,0,num.length-1);        if(num.length==1) return root;        //递归调用        arrayToBST(num,0,num.length-1,root);        return root;    }    private void arrayToBST(int[] num,int start,int end,TreeNode node){        if(start>end) return;        int middle=(start+end)>>1;        node.left=getMiddle(num,start,middle-1);        node.right=getMiddle(num,middle+1,end);        arrayToBST(num,start,middle-1,node.left);        arrayToBST(num,middle+1,end,node.right);    }    //获取中间的值作为节点    private TreeNode getMiddle(int[] num,int start,int end){        if(start>end) return null;        return new TreeNode(num[(start+end)>>1]);    }   }

【分析2-非原创】更简洁的写法。

https://oj.leetcode.com/discuss/6248/why-it-shows-runtime-error-when-using-my-recursive-code

 /**     * Definition for binary tree public class TreeNode { int val; TreeNode     * left; TreeNode right; TreeNode(int x) { val = x; } }     */    public class Solution {        public TreeNode sortedArrayToBST(int[] num) {            return arrayToBST(num, 0, num.length - 1);        }         private TreeNode arrayToBST(int[] num, int start, int end) {            if (start > end)                return null;            int middle = start + (end - start >> 1);            TreeNode root = new TreeNode(num[middle]);            if (start != end) {                root.left = arrayToBST(num, start, middle - 1);                root.right = arrayToBST(num, middle + 1, end);            }            return root;        }    }

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