【二叉树】二叉树前序、中序和后序遍历

前序遍历：

做题思路：用递归解决

1.加入递归结束条件：如果传入结点为NULL，直接return

2.将根结点本身传入数组

3.遍历左子树

4.遍历右子树

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

int* preorderTraversal(struct TreeNode* root, int* returnSize){
    int* ret = (int*)malloc(sizeof(int) * 100);
    *returnSize = 0;
    preOrderTraverse(root, ret, returnSize);
    return ret;
}

后序遍历：

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

int* postorderTraversal(struct TreeNode* root, int* returnSize){
    int* ret= (int*)malloc(sizeof(int) * 100);
    *returnSize = 0;
    postOrder(root, ret, returnSize);
    return ret;
}

中序遍历

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

int* inorderTraversal(struct TreeNode* root, int* returnSize){
    int* ret = (int*)malloc(sizeof(int) * 100);
    *returnSize = 0;
    inOrder(root, ret, returnSize);
    return ret;
}