Q4.6 find the ‘next’ node

Q: Write an algorithm to find the ‘next’ node (i.e., in-order successor) of a given node in a binary search tree where each node has a link to its parent.

A: 中序遍历的后继节点。 即找到最小的大于当前节点值的节点。

如果当前节点存在右孩子，那么结果就是当前节点右子树的最左节点。

如果当前节点没有右孩子，就向上寻找，node = node->parent, 一直到 node->parent == NULL 或者 node == node->parent->left,则node->parent就是结果

#include <iostream>
#include <cstring>
using namespace std;

const int maxn = 100;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode *parent;
    TreeNode () {}
 	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode *p, node[maxn];
int cnt;

void init(){
    p = NULL;
    memset(node, '\0', sizeof(node));
    cnt = 0;
}
void create_minimal_tree(TreeNode* &head, TreeNode *parent, int a[], int start, int end){
    if(start <= end){
        int mid = (start + end)>>1;
        node[cnt].val = a[mid];
        node[cnt].parent = parent;
        head = &node[cnt++];
        create_minimal_tree(head->left, head, a, start, mid-1);
        create_minimal_tree(head->right, head, a, mid+1, end);
    }
}

TreeNode *nextNode (TreeNode *node) {
	if (node->right) {
		TreeNode *cur = node->right;
		while (cur->left) {
			cur = cur->left;
		}
		return cur;
	} else {
		TreeNode *p = node->parent;
		while (p && node == p->right) {
			node = p;
			p = p->parent;
		}
		return p;
	}
}

int main(){
    int a[] = {
        0, 1, 2, 3, 4, 5, 6, 7, 8, 9
    };
    init();
    TreeNode *head = NULL;
    create_minimal_tree(head, NULL, a, 0, 9);
    cout<<"the head is "<<head->val<<endl;
    cout<<"the successor of head is: "<<(nextNode(head))->val<<endl;
    return 0;
}