题解 | #重建二叉树#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param preOrder int整型一维数组 
 * @param preOrderLen int preOrder数组长度
 * @param vinOrder int整型一维数组 
 * @param vinOrderLen int vinOrder数组长度
 * @return TreeNode类
 */
struct TreeNode* maketree(int pl , int pr , int* preOrder , int vl , int vr , int* vinOrder){
    if(pl > pr || pr < 0 || vl > vr || vr < 0) return NULL;
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = preOrder[pl];
    int i = vl;
    while(i <= vr){
        if(vinOrder[i] == preOrder[pl]) break;
        i++;
    }
    root->left = maketree(pl + 1 , pl + i - vl , preOrder , vl , i - 1 , vinOrder);
    root->right = maketree(pl + i - vl + 1 , pr , preOrder , i + 1 , vr , vinOrder);
    return root;
}
struct TreeNode* reConstructBinaryTree(int* preOrder, int preOrderLen, int* vinOrder, int vinOrderLen ) {
    // write code here
   int pl = 0 , pr = preOrderLen - 1;
   int vl = 0 , vr = vinOrderLen - 1;
   struct TreeNode* root = maketree(pl , pr , preOrder , vl , vr , vinOrder);
   return root;
}