Problem: 654. 最大二叉树

题目描述

思路

对于构造二叉树这类问题一般都是利用先、中、后序遍历，再将原始问题分解得出结果

1.定义递归函数build，每次将一个数组中的最大值作为当前子树的根节点构造二叉树；
2.每次找取当前范围内的最大值，作为当前的根节点；
3.递归求取出其左子树与右子树

复杂度

时间复杂度:

$O(n^2)$；其中n为二叉树节点的个数

空间复杂度:

$O(n)$

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    /**
     * Maximum Binary Tree
     *
     * @param nums Given array
     * @return TreeNode
     */
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        return build(nums, 0, nums.length - 1);
    }

    /**
     * Construction of binary tree function implementation
     *
     * @param nums Given array
     * @param low  Given the left endpoint of the array
     * @param high Given the right endpoint of the array
     * @return TreeNode
     */
    TreeNode build(int[] nums, int low, int high) {
        if (low > high) {
            return null;
        }
        int index = -1;
        int maxVal = Integer.MIN_VALUE;
        for (int i = low; i <= high; ++i) {
            if (maxVal < nums[i]) {
                maxVal = nums[i];
                index = i;
            }
        }
        //The root node is constructed first,
        // and then the left and right subtrees are constructed
        TreeNode root = new TreeNode(maxVal);
        root.left = build(nums, low, index - 1);
        root.right = build(nums, index + 1, high);
        return root;
    }
}