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二叉搜索树的后序遍历序列

题目:
输入一个整形数组。推断该数组是不是某二叉搜索树的后序遍历的结果.假设是则返回true,否则返回false. 假设输入的数组的随意两个数字都互不相同.

比如输入数组{5,7,6,9,11,10,8},则返回true.
{7,4,6,5}则返回false.

思路:
 后序遍历最后一个结点是根结点. 从第一个结点開始。找第一个大于根结点的结点,则这个结点的左边是左子树。右边是右字数.推断右子树有没有小于根结点的结点,假设有则不符合二叉搜索树的特性.若没有,则递归推断左右子树.

bool VerifySquenceOfBST(int arr[], int length)
{
    if (arr == NULL || length <= 0)
        return false;
    int root = *(arr+length-1);
    int i = 0;
    for (; i < length - 1; ++i)
    {
        if (arr[i] > root)
            break;
    }

    int j = i;

    for (; j < length - 1; ++j)
    {
        if (arr[j] < root)
            return false;
    }

    bool left = true;
    if (i > 0)
        left = VerifySquenceOfBST(arr,i);
    bool right = true;
    if (i < length - 1)
        right = VerifySquenceOfBST(arr + i, length - i - 1);

    return left&&right;
}

//相同的方法,我们也能够得到前序遍历序列
bool VerifyPreOrderOfBST(int arr[], int length)
{
    if (arr == NULL || length <= 0)
        return false;
    int root = arr[0];
    int i = 1;
    for (; i < length; ++i)
    {
        if (arr[i] > root)
            break;
    }
    int j = i;
//日了*了。这里把for写成了fot,提示我括号和分号不正确,我找了10几分钟眼睛都快
//花了!!

for (; j < length; ++j) { if (arr[j] < root) return false; } bool left = true; if (i > 1) left = VerifyPreOrderOfBST(arr+1, i); bool right = true; if (i < length) right = VerifyPreOrderOfBST(arr+i, length - i); return left&&right; } void Test() { printf("Test postOrder:\n"); int arr[] = {5,7,6,9,11,10,8}; bool res1 = VerifySquenceOfBST(arr,7); int arr1[] = {7,4,6,5}; bool res2 = VerifySquenceOfBST(arr1,7); bool res3 = VerifySquenceOfBST(NULL,1); printf("res1: %d\n",res1); printf("res2: %d\n",res2); printf("res3: %d\n",res3); printf("Test PreOrder:\n"); int arr2[] = {8,6,5,7,10,9,11}; bool res4 = VerifyPreOrderOfBST(arr2,7); int arr3[] = {8,6,5,10,9,7,11}; bool res5 = VerifyPreOrderOfBST(arr3,7); printf("res4: %d\n",res4); printf("res5: %d\n",res5); }

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二叉搜索树的后序遍历序列