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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
题目大意：给定一个升序序列的数组，将其转换为二叉搜索树。
解题思路：数组中间元素是根元素，根元素将数组划分为两部分，两个部分的中心元素分别为根元素的左右孩子，依次递推。。。可用迭代或递归来做。
解法一（迭代）：迭代需要自己额外记录下标
public TreeNode sortedArrayToBST(int[] num) { if (num == null || num.length == 0) { return null; } if (num.length == 1) { return new TreeNode(num[0]); } Deque<Pair> queue = new ArrayDeque<>(); TreeNode root = new TreeNode(num[(num.length - 1) / 2]); queue.add(new Pair(0, num.length - 1, root)); while (!queue.isEmpty()) { Pair pair = queue.poll(); int pre = pair.pre; int post = pair.post; int curr = (post + pre) / 2; TreeNode tNode = pair.node; if (curr > pre) { TreeNode left = new TreeNode(num[(curr - 1 + pre) / 2]); tNode.left = left; queue.add(new Pair(pre, curr - 1, left)); } if (post > curr) { TreeNode right = new TreeNode(num[(post + curr + 1) / 2]); tNode.right = right; queue.add(new Pair(curr + 1, post, right)); } } return root; } private static class Pair { int pre; int post; TreeNode node; public Pair(int pre, int post, TreeNode node) { this.pre = pre; this.post = post; this.node = node; } }
解法二（递归）：
public TreeNode sortedArrayToBST(int[] num) { if (num == null || num.length == 0) { return null; } TreeNode root = helper(0, num.length - 1, num); return root; } private TreeNode helper(int low, int high, int[] num) { if (low > high) { return null; } int mid = (low + high) / 2; TreeNode node = new TreeNode(num[mid]); node.left = helper(low, mid - 1, num); node.right = helper(mid + 1, high, num); return node; }
Convert Sorted Array to Binary Search Tree——LeetCode
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原文地址：http://www.cnblogs.com/aboutblank/p/4442715.html
