题目介绍

题目分析

题目要求我们输出二叉树按前序遍历排列的每个节点的值。

解题思路

1.创建一个数组来储存二叉树节点的值

2.根据二叉树的大小来开辟数组的大小

int Treesize(struct TreeNode* root)
{if(root==NULL)return 0;return 1+Treesize(root->left)+Treesize(root->right);
}int*new=(int*)malloc(sizeof(int)*Treesize(root));

3.边前序遍历边向创建的数组中存入二叉树节点的值

void preorder(struct TreeNode* root,int*new,int*newsize)
{if(root==NULL)return;new[(*newsize)++]=root->val;preorder(root->left,new,newsize);preorder(root->right,new,newsize);
}

完整代码

int Treesize(struct TreeNode* root)
{if(root==NULL)return 0;return 1+Treesize(root->left)+Treesize(root->right);
}void preorder(struct TreeNode* root,int*new,int*newsize)
{if(root==NULL)return;new[(*newsize)++]=root->val;preorder(root->left,new,newsize);preorder(root->right,new,newsize);
}
int* preorderTraversal(struct TreeNode* root, int* returnSize) {int*new=(int*)malloc(sizeof(int)*Treesize(root));*returnSize=0;preorder(root,new,returnSize);return new;
}