LeeCode刷题 二叉树 C语言

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LeeCode二叉树
刚开始刷题没多久,我会陆续补充

二叉树的遍历

前序遍历

题目
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代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
#define MAX 1024

void preorder(struct TreeNode* root, int *arr, int* index){
    if(!root) return;
    arr[(*index)++] = root->val;
    preorder(root->left, arr, index);
    preorder(root->right, arr, index);
}
int* preorderTraversal(struct TreeNode* root, int* returnSize){
    *returnSize = 0;
    int *arr = (int*)malloc(MAX * sizeof(int));
    if(root!=NULL){
    	preorder(root, arr, returnSize);
	}
    return arr;
}

结果
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中序遍历

题目
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代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
#define MAXSIZE 1024

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
void inorder(struct TreeNode* root, int* ret, int* index ){
    if(root != NULL){
        inorder(root->left, ret, index);
        ret[(*index)++] = root->val;
        inorder(root->right, ret, index);
    }
}  
    
int* inorderTraversal(struct TreeNode* root, int* returnSize){
    int *ret = (int*)malloc(sizeof(int)*MAXSIZE);//分配存储空间
    *returnSize = 0;
    if(root != NULL){
       inorder(root, ret, returnSize);
    } 
    return ret;
}

结果:
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后序遍历

题目
在这里插入图片描述
代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
#define MAXSIZE 1024

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
void postorder(struct TreeNode* root, int *ret, int *index){
    if(root!=NULL){
        postorder(root->left, ret, index);
        postorder(root->right, ret, index);
        ret[(*index)++] = root->val;
    }
}
int* postorderTraversal(struct TreeNode* root, int* returnSize){
    int *ret = (int*)malloc(sizeof(int)*MAXSIZE);
    *returnSize = 0;
    if(root != NULL){
        postorder(root, ret, returnSize);
    }
    return ret;
}

结果:
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层序遍历

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转载自blog.csdn.net/weixin_45468845/article/details/107184573