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construct a binary tree from inorder and dfs

Vivacious Vendace answered on May 12, 2022 Popularity 3/10 Helpfulness 1/10

answer construct a binary tree from inorder and dfs

construct a binary tree from inorder and dfs

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#include <bits/stdc++.h>

using namespace std;

typedef struct node
{
    int data;
    struct node *left, *right;
} Node;

int iDFS = 0;

Node *createNode(int data)
{
    Node *root = (Node *)malloc(sizeof(struct node));
    root->data = data;
    root->left = NULL;
    root->right = NULL;
    return root;
}

Node *createTree(Node *root, int *in, int *dfs, int istart, int iend)
{
    if (istart > iend)
    {
        return NULL;
    }

    if (istart == iend)
    {
        int rootVal = dfs[iDFS];
        root = createNode(rootVal);
        iDFS++;
        return root;
    }
    int rootVal = dfs[iDFS];
    root = createNode(rootVal);
    int ele = -1;
    for (int i = istart; i <= iend; i++)
    {
        if (dfs[iDFS] == in[i])
            ele = i;
    }
    iDFS++;
    if (ele != -1)
    {
        root->left = createTree(root->left, in, dfs, istart, ele - 1);
        root->right = createTree(root->right, in, dfs, ele + 1, iend);
    }
    return root;
}

void inorderTraversal(struct node *root)
{
    if (root != NULL)
    {
        inorderTraversal(root->left);
        cout << root->data << " ";
        inorderTraversal(root->right);
    }
}

int main()
{
    int in[] = {8, 4, 2, 5, 1, 6, 3, 7, 9};
    int dfs[] = {1, 2, 4, 8, 5, 3, 6, 7, 9};
    int n = sizeof(in) / sizeof(in[0]) - 1;
    Node *root = createTree(root, in, dfs, 0, n);
    inorderTraversal(root);
    return 0;
}

Popularity 3/10 Helpfulness 1/10 Language whatever

Source: Grepper

Tags: binary-tree construct inorder whatever

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Contributed on May 12 2022

Vivacious Vendace

0 Answers Avg Quality 2/10

construct a binary tree from inorder and dfs

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