删除二叉树中的度数为1的所有结点

来源：互联网发布：淘宝供销平台一件代发编辑：程序博客网时间：2024/04/25 09:08

要求：结点删除后其唯一的子节点代替它的位置。

这里用到了递归的方法，不断遍历每个节点的左右子树，将度数为一的结点删除。

#include <iostream>using namespace std;struct Node{int v;Node* left;Node* right;};void print_div(int p){for(int i=0; i < p; i++){cout<<"-";}}void print_tree(Node* node, int p){if(node != NULL){print_div(p);cout<<node->v<<endl;if(node->left != NULL || node->right != NULL){print_tree(node->left, p+1);print_tree(node->right, p+1);}}else{print_div(p);cout<<endl;}}void print_tree(Node* node){print_tree(node, 0);}void delete_one_degree_node(Node*& node){if(node == NULL)return;if(node->left != NULL && node->right == NULL) //只有左子树{node = node->left;delete_one_degree_node(node);}else if(node->right != NULL && node->left == NULL)  //只有右子树{node = node->right;delete_one_degree_node(node);}else if(node->left != NULL && node->right != NULL) //左右子树均存在{delete_one_degree_node(node->left);delete_one_degree_node(node->right);}}int main(){Node n[10] = {0};Node* tree = n;for(int i=0; i < 10; i++){n[i].v = i;}tree[0].left = &tree[1];tree[0].right = &tree[2];tree[1].left = &tree[3];tree[1].right = &tree[4];tree[2].left = NULL;tree[2].right = &tree[6];tree[3].left = &tree[7];tree[3].right = &tree[8];tree[4].left = &tree[9];cout<<"Original:"<<endl;print_tree(tree);delete_one_degree_node(tree);cout<<"After:"<<endl;print_tree(tree);return 0;}

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