TOP101题解 | BM25#二叉树的后序遍历#
二叉树的后序遍历
https://www.nowcoder.com/practice/1291064f4d5d4bdeaefbf0dd47d78541
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * @author Senky * @date 2023.05.08 * @par url https://www.nowcoder.com/creation/manager/content/584337070?type=column&status=-1 * @brief * @param root TreeNode类 * @return int整型一维数组 * @return int* returnSize 返回数组行数 */ int str[]; void PostorderVisit(struct TreeNode* root,int *returnSize) { if(root) { PostorderVisit(root->left,returnSize); PostorderVisit(root->right,returnSize); str[(*returnSize)++] = root->val; } } int* postorderTraversal(struct TreeNode* root, int* returnSize ) { // write code here PostorderVisit(root,returnSize); return str; }#TOP101#
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系列的题解