算法之二叉树的三种遍历

// 前序遍历
void preorder(struct TreeNode* root, int *ret, int* returnSize)
{
    if(!root)
    {
        return;
    }
    ret[(*returnSize)++] = root->val;
    preorder(root->left, ret, returnSize);
    preorder(root->right, ret, returnSize);
}

int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    int *ret=(int *)malloc(sizeof(int) * 2000);
    *returnSize = 0;
    preorder(root, ret, returnSize);
    return ret;
}

leetcode94
中序遍历

/**
 * 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().
 */

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

int* inorderTraversal(struct TreeNode* root, int* returnSize) {
       int *ret=(int *)malloc(sizeof(int) * 2000);
    *returnSize = 0;
    preorder(root, ret, returnSize);
    return ret; 
}

后序遍历

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

int* postorderTraversal(struct TreeNode* root, int* returnSize) {
    int *ret=(int *)malloc(sizeof(int) * 2000);
    *returnSize = 0;
    preorder(root, ret, returnSize);
    return ret;
}