二叉树公共祖先

// 将两个保存祖先的公共部门组合并成一个
// 合并后的祖先,顺序是反的,就是最后一个才是最近的祖先
void mergeCommon(char stack[],  char stack1[], char stack2[], int top1, int top2) {
    for (int i = 0; i <= top1 && i <= top2; i++) {
        if (stack1[i] == stack2[i]) {
            stack[i] = stack1[i];
        } else {
            break;
        }
    }
}

void copyStack(BTNode *stack1[], char stack2[], int top) {
    for (int i = 0; i <= top; i++) {
        stack2[i] = stack1[i]->data;
     }
}

// 寻找两个节点的共同祖先 思路,用后序非递归算法
// 用两个辅助栈,一个栈保存
void printCommonAncestor(BTNode *root, char y, char z) {
    if (root != nullptr) {
        int tag[20]; // 标记每个节点的右子树是否被访问过
        BTNode *stack[maxsize];
        char stacky[20], stackz[20]; // 辅助栈,保留节点y的祖先和节点z的祖先
        int topy = -1, topz = -1;
        int top = -1;
        BTNode *q;
        q = root;
        while (q != nullptr || top != -1) {
            while (q != nullptr) {
                stack[++top] = q;
                q = q->left;
                tag[top] = 0;
            }
            while (top != -1 && tag[top] == 1) {
                q = stack[top--];
                if (q->data == y) {
                    // 将栈里面剩余的复制到stacky里面
                    copyStack(stack, stacky, top);
                    topy = top;
                }
                if (q->data == z) {
                    copyStack(stack, stackz, top);
                    topz = top;
                }
            }
            if (top != -1) {
                tag[top] = 1;
                q = stack[top];
                q = q->right;
            } else {
                break;
            }
        }

        // 最后stacky 和 stackz里面保存的都是各自的祖先
        char commonAncestor[maxsize];
        mergeCommon(commonAncestor, stacky, stackz, topy, topz);
        printf("");
    }
}

猜你喜欢

转载自my.oschina.net/u/3139515/blog/1577918
今日推荐