Construct Binary Tree from Preorder and Inorder Traversal Solutions in C++

// Source : https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
// Author : Hao Chen
// Date   : 2014-07-09



#include <stdio.h>
#include <vector>
#include <queue>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode *buildTree(vector<int>& preorder, int& preidx, vector<int>& inorder);

TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
    int preidx=0;
    return buildTree(preorder, preidx, inorder);
}

TreeNode *buildTree(vector<int>& preorder, int& preidx, vector<int>& inorder) {

    if (preorder.size()<=0 || inorder.size()<=0 ) return NULL;

    TreeNode *root = new TreeNode(preorder[preidx]);
    if (inorder.size()==1){
        return root;
    }

    int i;
    for(i=0; i<inorder.size(); i++){
        if (inorder[i] == preorder[preidx]){
            break;
        }
    }

    //error: not found
    if (i == inorder.size()) return NULL;

    if (preidx+1 >= preorder.size()){
        return root;
    }

    
    vector<int> v(inorder.begin(), inorder.begin()+i);
    if (v.size()>0) {
        preidx++;
        root->left = buildTree(preorder, preidx, v);
    }

    v.clear();
    v.assign(inorder.begin()+i+1, inorder.end());
    if (v.size()>0) {
        preidx++;
        root->right = buildTree(preorder, preidx, v);
    }

    return root;
}

void printTree_pre_order(TreeNode *root)
{
    if (root == NULL){
        printf("# ");
        return;
    }
    printf("%c ", root->val );

    printTree_pre_order(root->left);
    printTree_pre_order(root->right);
}

void printTree_in_order(TreeNode *root)
{
    if (root == NULL){
        printf("# ");
        return;
    }

    printTree_in_order(root->left);
    printf("%c ", root->val );
    printTree_in_order(root->right);
}


void printTree_level_order(TreeNode *root)
{
    queue<TreeNode*> q;
    q.push(root);
    while (q.size()>0){
        TreeNode* n = q.front();
        q.pop();
        if (n==NULL){
            printf("# ");
            continue;
        }
        printf("%c ", n->val);
        q.push(n->left);
        q.push(n->right);
    }
    printf("\n");
}


int main()
{
    int pre_order[]={'F', 'B', 'A', 'D', 'C', 'E', 'G', 'I', 'H'};
    int  in_order[]={'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I'};
    vector<int> preorder( pre_order, pre_order + 9 );
    vector<int>  inorder(  in_order,  in_order + 9 );

    TreeNode* tree = buildTree(preorder, inorder);

    printTree_level_order(tree);
    printTree_pre_order(tree);
    printf("\n");
    printTree_in_order(tree);
    printf("\n");
    
    return 0;
}

#include <vector>
#include <cstddef>
#include <algorithm>

using std::vector; using std::next; using std::prev;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    using cIter = vector<int>::const_iterator;
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        return buildTree(preorder.cbegin(), preorder.cend(), inorder.cbegin(), inorder.cend());
    }
    TreeNode *buildTree(cIter preBeg, cIter preEnd, cIter inBeg, cIter inEnd) {
        if (preBeg >= preEnd || inBeg >= inEnd) return NULL;
        TreeNode *root = new TreeNode(*preBeg);
        auto inRoot = std::find(inBeg, inEnd, root->val);
        root->left = buildTree(next(preBeg), next(preBeg, inRoot-inBeg+1), inBeg, inRoot);
        root->right = buildTree(next(preBeg, inRoot-inBeg+1), preEnd, next(inRoot), inEnd);
        return root;
    }
};

/// Source : https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/
/// Author : liuyubobobo
/// Time   : 2018-06-03

#include <iostream>
#include <vector>
#include <cassert>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

/// Recursive
/// Time Complexity: O(n*h) where n is the num of node in th tree
///                         and h is the height of the tree
/// Space Complexity: O(h)
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return buildTree(preorder, 0, preorder.size(), inorder, 0, inorder.size());
    }

private:
    TreeNode* buildTree(const vector<int>& preorder, int preorderL, int preorderR,
                        const vector<int>& inorder, int inorderL, int inorderR){

        if(inorderL >= inorderR){
            assert(preorderL >= preorderR);
            return NULL;
        }

        if(inorderL + 1 == inorderR){
            assert(preorderL + 1 == preorderR);
            return new TreeNode(inorder[inorderL]);
        }

        TreeNode* root = new TreeNode(preorder[preorderL]);
        int rootPos = find(inorder.begin() + inorderL, inorder.begin() + inorderR, root->val) - inorder.begin();
        assert(rootPos >= inorderL && rootPos < inorderR);

        int lsize = rootPos - inorderL;
        int rsize = inorderR - (rootPos + 1);
        root->left = buildTree(preorder, preorderL + 1, preorderL + 1 + lsize, inorder, inorderL, rootPos);
        root->right = buildTree(preorder, preorderL + 1 + lsize, preorderR, inorder, rootPos + 1, inorderR);
        return root;
    }
};


int main() {

    return 0;
}