106. Construct Binary Tree from Inorder and Postorder Traversal

Description

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

 

Example 1:

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

Example 2:

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

 

Constraints:

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

Solutions

Solution 1: Hash Table + Recursion

The last node in the post-order traversal is the root node. We can find the position of the root node in the in-order traversal, and then recursively construct the left and right subtrees.

Specifically, we first use a hash table $d$ to store the position of each node in the in-order traversal. Then we design a recursive function $dfs(i, j, n)$, where $i$ and $j$ represent the starting positions of the in-order and post-order traversals, respectively, and $n$ represents the number of nodes in the subtree. The function logic is as follows:

• If $n \leq 0$, it means the subtree is empty, return a null node.
• Otherwise, take out the last node $v$ of the post-order traversal, and then find the position $k$ of $v$ in the in-order traversal using the hash table $d$. Then the number of nodes in the left subtree is $k - i$, and the number of nodes in the right subtree is $n - k + i - 1$.
• Recursively construct the left subtree $dfs(i, j, k - i)$ and the right subtree $dfs(k + 1, j + k - i, n - k + i - 1)$, connect them to the root node, and finally return the root node.

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++GoTypeScriptRust

# 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, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        def dfs(i: int, j: int, n: int) -> Optional[TreeNode]:
            if n <= 0:
                return None
            v = postorder[j + n - 1]
            k = d[v]
            l = dfs(i, j, k - i)
            r = dfs(k + 1, j + k - i, n - k + i - 1)
            return TreeNode(v, l, r)

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

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

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        this.postorder = postorder;
        int n = inorder.length;
        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 = postorder[j + n - 1];
        int k = d.get(v);
        TreeNode l = dfs(i, j, k - i);
        TreeNode r = dfs(k + 1, j + k - i, n - k + i - 1);
        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>& inorder, vector<int>& postorder) {
        unordered_map<int, int> d;
        int n = inorder.size();
        for (int i = 0; i < n; ++i) {
            d[inorder[i]] = i;
        }
        function<TreeNode*(int, int, int)> dfs = [&](int i, int j, int n) -> TreeNode* {
            if (n <= 0) {
                return nullptr;
            }
            int v = postorder[j + n - 1];
            int k = d[v];
            auto l = dfs(i, j, k - i);
            auto r = dfs(k + 1, j + k - i,