105. Construct Binary Tree from Preorder and Inorder Traversal

Description

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

 

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

 

Constraints:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder and inorder consist of unique values.
• Each value of inorder also appears in preorder.
• preorder is guaranteed to be the preorder traversal of the tree.
• inorder is guaranteed to be the inorder traversal of the tree.

Solutions

Solution 1: Hash Table + Recursion

The first node $preorder[0]$ in the pre-order sequence is the root node. We find the position $k$ of the root node in the in-order sequence, which can divide the in-order sequence into the left subtree $inorder[0..k]$ and the right subtree $inorder[k+1..]$.

Through the intervals of the left and right subtrees, we can calculate the number of nodes in the left and right subtrees, assumed to be $a$ and $b$. Then in the pre-order nodes, the $a$ nodes after the root node are the left subtree, and the $b$ nodes after that are the right subtree.

Therefore, we design a function $dfs(i, j, n)$, where $i$ and $j$ represent the starting positions of the pre-order sequence and the in-order sequence, respectively, and $n$ represents the number of nodes. The return value of the function is the binary tree constructed with $preorder[i..i+n-1]$ as the pre-order sequence and $inorder[j..j+n-1]$ as the in-order sequence.

The execution process of the function $dfs(i, j, n)$ is as follows:

• If $n \leq 0$, it means there are no nodes, return a null node.
• Take out the first node $v = preorder[i]$ of the pre-order sequence as the root node, and then use the hash table $d$ to find the position $k$ of the root node in the in-order sequence. Then the number of nodes in the left subtree is $k - j$, and the number of nodes in the right subtree is $n - k + j - 1$.
• Recursively construct the left subtree $l = dfs(i + 1, j, k - j)$ and the right subtree $r = dfs(i + 1 + k - j, k + 1, n - k + j - 1)$.
• Finally, return the binary tree with $v$ as the root node and $l$ and $r$ as the left and right subtrees, respectively.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

Python3JavaC++GoTypeScriptRustJavaScript

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        def dfs(i: int, j: int, n: int) -> Optional[TreeNode]:
            if n <= 0:
                return None
            v = preorder[i]
            k = d[v]
            l = dfs(i + 1, j, k - j)
            r = dfs(i + 1 + k - j, k + 1, n - k + j - 1)
            return TreeNode(v, l, r)

        d = {v: i for i, v in enumerate(inorder)}
        return dfs(0, 0, len(preorder))

/**
 * 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 {
    private int[] preorder;
    private Map<Integer, Integer> d = new HashMap<>();

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int n = preorder.length;
        this.preorder = preorder;
        for (int i = 0; i < n; ++i) {
            d.put(inorder[i], i);
        }
        return dfs(0, 0, n);
    }

    private TreeNode dfs(int i, int j, int n) {
        if (n <= 0) {
            return null;
        }
        int v = preorder[i];
        int k = d.get(v);
        TreeNode l = dfs(i + 1, j, k - j);
        TreeNode r = dfs(i + 1 + k - j, k + 1, n - 1 - (k - j));
        return new TreeNode(v, l, r);
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int n = preorder.size();
        unordered_map<int, int> d;
        for (int i = 0; i < n; ++i) {
            d[inorder[i]] = i;