# 105.Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal

## 105. Construct Binary Tree from Preorder and Inorder Traversal

## 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
```

## 代码

### Approach 1: Recursion

inorder数组用来获取反向索引

preorder数组用来递归

Time complexity : O(*N*) & Space complexity : O(*N*), since we store the entire tree.

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

        return dfs(0, len - 1);
    }

    private TreeNode dfs(int start, int end) {
        if (start > end)      return null;

        int rootVal = preorder[preIndex];
        TreeNode root = new TreeNode(rootVal);
        int index = map.get(rootVal);
        preIndex++;
        root.left = dfs(start, index - 1);
        root.right = dfs(index + 1, end);

        return root;
    }
}
```


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