二叉搜索树找前驱和后继

输入N个数,找每个数的前驱和后继。如果没有前驱或后继,输出-1;

思路:

如果有右子树,则右子树的最小值为当前节点的后继;否则后继为当前节点往祖先搜索,第一次是左孩子的节点的父亲的值;

如果有左子树,则左子树的最大值为当前节点的前驱;否则前驱为当前节点往祖先搜索,第一次是右孩子的节点的父亲的值;

#include "bits/stdc++.h"
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
struct BST {
    int value;
    BST* father;
    BST* lson;
    BST* rson;
}* root;
BST* init(int val) {
    BST* point = (BST*)malloc(sizeof(BST));
    point->value = val;
    point->lson = point->rson = NULL;
    return point;
}
void insert(int val) {
    BST* father = NULL;
    BST* now = root;
    while (now != NULL) {
        if (now->value == val) {
            return;
        }
        father = now;
        if (now->value < val) {
            now = now->rson;
        } else {
            now = now->lson;
        }
    }
    if (father == NULL) {
        root = init(val);
        root->father = NULL;
    } else if (father->value < val) {
        father->rson = init(val);
        father->rson->father = father;
    } else {
        father->lson = init(val);
        father->lson->father = father;
    }
}
int minOfBST(BST* point) {
    while (point->lson != NULL) {
        point = point->lson;
    }
    return point->value;
}
int maxOfBST(BST* point) {
    while (point->rson != NULL) {
        point = point->rson;
    }
    return point->value;
}
int precursor (int val) {
    BST* point = root;
    while (point != NULL) {
        if (point->value == val) {
            if (point->lson != NULL) {
                return maxOfBST(point->lson);
            } else {
                while (point->father != NULL && point->father->rson != point) {
                    point = point->father;
                }
                if (point->father == NULL){
                    return -1;
                } else {
                    return point->father->value;
                }
            }
        }
        if (val > point->value) {
            point = point->rson;
        } else {
            point = point->lson;
        }
    }
}
int successor (int val) {
    BST* point = root;
    while (point != NULL) {
        if (point->value == val) {
            if (point->rson != NULL) {
                return minOfBST(point->rson);
            } else {
                while (point->father != NULL && point->father->lson != point) {
                    point = point->father;
                }
                if (point->father == NULL){
                    return -1;
                } else {
                    return point->father->value;
                }
            }
        }
        if (val > point->value) {
            point = point->rson;
        } else {
            point = point->lson;
        }
    }
}
int main() {
    int N, val;
    queue<int>q;
    scanf("%d", &N);
    while (N--) {
        scanf("%d", &val);
        insert(val);
        q.push(val);
    }
    while (!q.empty()) {
        printf("%d ", precursor(q.front()));
        printf("%d\n", successor(q.front()));
        q.pop();
    }
    return 0;
}

 

posted @ 2019-01-04 21:46  Jathon-cnblogs  阅读(791)  评论(0编辑  收藏  举报