【LeetCode】 105. Construct Binary Tree from Preorder and Inorder Traversal 从前序与中序遍历序列构造二叉树（Medium）（JAVA）

题目地址: https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

题目描述：

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

题目大意

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

解题方法

1、前序遍历的根节点在第一个；中序遍历的根节点在中间，找到根节点，就能分为左右子树
2、迭代更新完整的树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            map.put(inorder[i], i);
        }
        return bH(preorder, inorder, map, 0, preorder.length - 1, 0, inorder.length - 1);
    }

    public TreeNode bH(int[] preorder, int[] inorder, Map<Integer, Integer> map, int pStart, int pEnd, int iStart, int iEnd) {
        if (pStart > pEnd || iStart > iEnd) return null;
        int mid = map.get(preorder[pStart]);
        TreeNode root = new TreeNode(preorder[pStart]);
        root.left = bH(preorder, inorder, map, pStart + 1, pStart + mid - iStart, iStart, mid - 1);
        root.right = bH(preorder, inorder, map, pStart + mid - iStart + 1, pEnd, mid + 1, iEnd);
        return root;
    }
}

执行用时 : 3 ms, 在所有 Java 提交中击败了 81.03% 的用户
内存消耗 : 40.1 MB, 在所有 Java 提交中击败了 66.67% 的用户