树形数据结构算法

Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:

1. The root is the maximum number in the array.
2. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
3. The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.

Construct the maximum tree by the given array and output the root node of this tree.

Example 1:

Input: [3,2,1,6,0,5]Output: return the tree root node representing the following tree:      6    /   \   3     5    \    /      2  0          \        1

Note:

1. The size of the given array will be in the range [1,1000].

以上是LeetCode的一道关于二叉树的算法题，之前没怎么刷过树形数据结构的题，今天来总结一下这样的题

public class Solution {    public TreeNode constructMaximumBinaryTree(int[] nums) {        if (nums == null) return null;        return build(nums, 0, nums.length - 1);    }        private TreeNode build(int[] nums, int start, int end) {        if (start > end) return null;                int idxMax = start;        for (int i = start + 1; i <= end; i++) {            if (nums[i] > nums[idxMax]) {                idxMax = i;            }        }                TreeNode root = new TreeNode(nums[idxMax]);                root.left = build(nums, start, idxMax - 1);        root.right = build(nums, idxMax + 1, end);                return root;    }}

这是直接粘贴人家大神的算法。。。当做样本以此来学习一下。

其实总体的思路是：一直递归，找根节点；包括树根结点和左右子树的根节点。

 int idxMax = start;        for (int i = start + 1; i <= end; i++) {            if (nums[i] > nums[idxMax]) {                idxMax = i;            }        }

这段代码是找最大值的下标，用TreeNode root = new TreeNode（nums[idxMax]）存储最大值

root.left = build(nums,start,idxMax-1)   递归建立左子树，root.right = build(nums,idxMax+1,end)  递归建立右子树

这样的题最主要的是如何建立递归模型。