c语言刷 DFS题记录

144. 二叉树的前序遍历

/**
 * 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().
 */

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

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

145. 二叉树的后序遍历

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

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

105. 从前序与中序遍历序列构造二叉树

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

struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize)
{
    if (preorder == NULL || inorder == NULL) {
        return NULL;
    }
    if (preorderSize == 0 || inorderSize == 0) {
        return NULL;
    }
    if (preorderSize != inorderSize) {
        return NULL;
    }

    struct TreeNode *res = (struct TreeNode *)malloc(sizeof(struct TreeNode));
    int rootVal = preorder[0];
    int leftLen = 0;
    for (int i = 0; i < inorderSize; i++) {
        if (rootVal == inorder[i]) {
            leftLen = i;
            break;
        }
    }
    int rightLen = inorderSize - leftLen - 1;

    res->val = rootVal;
    res->left = buildTree(&preorder[1], leftLen, inorder, leftLen);
    res->right = buildTree(&preorder[leftLen + 1], rightLen, &inorder[leftLen + 1], rightLen);
    return res;
}

106. 从中序与后序遍历序列构造二叉树

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

struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize)
{
    if (inorder == NULL || postorder == NULL) {
        return NULL;
    }
    if (inorderSize == 0 || postorderSize == 0) {
        return NULL;
    }
    if (inorderSize != postorderSize) {
        return NULL;
    }

    struct TreeNode* res = (struct TreeNode *)malloc(sizeof(struct TreeNode));
    int rootVal = postorder[postorderSize - 1];
    int leftNum = 0;
    int rightNum = 0;
    for (int i = 0; i < inorderSize; i++) {
        if (rootVal == inorder[i]) {
            leftNum = i;
            break;
        }
    }
    rightNum = inorderSize - leftNum - 1;
    res->val = rootVal;
    res->left = buildTree(inorder, leftNum, postorder, leftNum);
    res->right = buildTree(&inorder[leftNum + 1], rightNum, &postorder[leftNum], rightNum);
    return res;
}