leetcode [Convert Sorted Array to Binary Search Tree]//待整理多种解法
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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ /* 本题要求求出给定有序数列的高度平衡二叉查找树,但是由二叉查找树的构造定理构造出的树就已经是高度平衡的了 所以只需按二叉查找树的构造定理来进行构造即可,构造过程实际是一个DFS的过程,先根结点,然后左子结点,然后右子结点 */ public class Solution { /*public TreeNode sortedArrayToBST(int[] nums) { TreeNode root = null; int low = 0, high = nums.length - 1; recurse(root, low, high, nums); return root; } public void recurse(TreeNode root, int low, int high, int[] nums){ //编写递归时只用考虑根结点的情况,因为左右子结点是用来递归调用的 //递归出口 if(low > high) return; int mid = (low + high) / 2; root = new TreeNode(nums[mid]);//只处理根结点 //递归 recurse(root.left, low, mid - 1, nums); recurse(root.right, mid + 1, high, nums); }*///注释部分不行,又涉及到了Java中只是值传递,//即使传引用传的也只是引用地址的拷贝,root = new TreeNode(x)后传入方法的root已经被覆盖了,与方法外的引用无关了//所以下面的写法中用返回值来,返回值返回的就是一个已经建好的树的根结点 public TreeNode sortedArrayToBST(int[] nums) { TreeNode root = null; int low = 0, high = nums.length - 1; root = recurse(low, high, nums); return root; } public TreeNode recurse(int low, int high, int[] nums){ TreeNode root = null; //编写递归时只用考虑根结点的情况,因为左右子结点是用来递归调用的 //递归出口 if(low > high) return null; int mid = (low + high) / 2; root = new TreeNode(nums[mid]);//只处理根结点 //递归 root.left = recurse(low, mid - 1, nums); root.right = recurse(mid + 1, high, nums); return root; }}
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