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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
知识点:
二叉搜素树:
1. 如果root有左子节点,则所有左边的节点<root;
2. 如果root有右子节点,则所有右边的节点>root;
3. root.left也是二叉搜索树
4. root.right也是二叉搜索树
题目要求这个二叉搜索树是平衡的。
思路:
设middle = (left+middle)/2,left是起始下标,right是结束下标,该问题的解释solve(num, left,right)
1. 根节点root = num[middle];
2. 根节点的左子结点root.left = solve(num, left, middle-1);
3. 根节点的右子节点root.right = solve(num, middle+1, right);
4. 递归调用以上过程
代码如下:
public TreeNode sortedArrayToBST(int[] num) { if(num == null || num.length == 0) return null; return sortedArrayToBST(num, 0, num.length-1); } public TreeNode sortedArrayToBST(int[] num, int start, int over) { if(start == over) return new TreeNode(num[start]); if(start > over) return null; int middle = (start+over)/2; TreeNode root = new TreeNode(num[middle]); root.left = sortedArrayToBST(num, start, middle-1); root.right = sortedArrayToBST(num, middle+1, over); return root; }
LeetCode-108 Convert Sorted Array to Binary Search Tree
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原文地址:http://www.cnblogs.com/linxiong/p/4324399.html