从前序与中序遍历序列构造二叉树

题目

题目来源 : LeetCode

实现

struct TreeNode* _buildTree(int* preorder, int* inorder, int* curindex, int inbegin, int inend){
    //创建根
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = preorder[*curindex];
    //在中序遍历中查找根的位置
    int rootindex = inbegin;
    for(; rootindex <= inend; ++rootindex){
        if(inorder[rootindex] == root->val){
            break;
        }
    }
    //rootindex 中序遍历根的位置
    //创建左子树
    if(inbegin < rootindex){
        ++(*curindex);
        root->left = _buildTree(preorder, inorder, curindex, inbegin, rootindex - 1);
    }else{
        root->left = NULL;
    }
    //创建右子树
    if(inend > rootindex){
        ++(*curindex);
        root->right = _buildTree(preorder, inorder, curindex, rootindex + 1, inend);
    }else{
        root->right = NULL;
    }
    return root;
}

struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize){
    int curindex = 0;
    if(preorderSize == 0 || inorderSize == 0){
        return NULL;
    }
    return _buildTree(preorder, inorder, &curindex, 0, inorderSize - 1);
}