sol 105, 106, 654 and 704

Author: ioMayday

c_code/105_buildTree_pre_in.c

@@ -0,0 +1,44 @@
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     struct TreeNode *left;
+ *     struct TreeNode *right;
+ * };
+ */
+
+// Good Job，理解到位，一次写对！！！
+// 根据前序和中序，构建一个二叉树
+struct TreeNode* build(int* preorder, int preStart, int preEnd, int* inorder, int inStart, int inEnd)
+{
+    // base case
+    if (inStart > inEnd) return NULL; // 叶子节点
+
+    // root operation
+    struct TreeNode* root = (struct TreeNode *)malloc(sizeof(struct TreeNode));
+    // 前序遍历结果
+    root->val = preorder[preStart];
+    // 中序遍历来分界
+    int i, idx = -1;
+    for (i = inStart; i <= inEnd; i++) {
+        if(inorder[i] == root->val) {
+            idx = i;
+            break;
+        }
+    }
+
+    // 递归生成左右子树
+    int leftSize = idx - inStart;
+    root->left = build(preorder, preStart + 1, preStart + leftSize,
+                      inorder, inStart, idx - 1);
+    root->right = build(preorder, preStart + leftSize + 1, preEnd,
+                      inorder, idx + 1, inEnd);
+
+    return root;
+}
+
+struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize){
+    if (preorder == NULL || inorder == NULL) return NULL;
+
+    return build(preorder, 0, preorderSize - 1, inorder, 0, inorderSize - 1);
+}

c_code/106_buildTree_post_in.c

@@ -0,0 +1,46 @@
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     struct TreeNode *left;
+ *     struct TreeNode *right;
+ * };
+ */
+ 
+// Good Job，理解到位，一次写对！！！
+// 根据后序和中序，构建一个二叉树
+// 在上个105题基础上，改写下标，8min搞定
+
+struct TreeNode* build(int* inorder, int inStart, int inEnd, int* postorder, int postStart, int postEnd)
+{
+    // base case
+    if (inStart > inEnd) return NULL; // 叶子节点
+
+    // root operation
+    struct TreeNode* root = (struct TreeNode *)malloc(sizeof(struct TreeNode));
+    // 后序遍历结果
+    root->val = postorder[postEnd];
+    // 中序遍历来分界
+    int i, idx = -1;
+    for (i = inStart; i <= inEnd; i++) {
+        if(inorder[i] == root->val) {
+            idx = i;
+            break;
+        }
+    }
+
+    // 递归生成左右子树
+    int leftSize = idx - inStart;
+    root->left = build(inorder, inStart, idx - 1,
+                      postorder, postStart, postStart + leftSize - 1);
+    root->right = build(inorder, idx + 1, inEnd,
+                      postorder, postStart + leftSize, postEnd - 1);
+
+    return root;
+}
+
+struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize){
+    if (postorder == NULL || inorder == NULL) return NULL;
+
+    return build(inorder, 0, inorderSize - 1, postorder, 0, postorderSize - 1);
+}

c_code/654_constructMaximumBinaryTree.c

@@ -0,0 +1,47 @@
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     struct TreeNode *left;
+ *     struct TreeNode *right;
+ * };
+ */
+
+ // 闭区间 [left, right]
+struct TreeNode* buildMaxTree(int* nums, int left, int right) 
+{
+    // base case
+    if (left > right) return NULL;
+
+    struct TreeNode* root = (struct TreeNode *)malloc(sizeof(struct TreeNode));
+    // 找到最大值做当前的根节点
+	// debug code
+    // int i, idx, maxVal = nums[left];
+    // for (i = left + 1; i <= right; i++) { // 会报内存越界，待查看bug;  =>需同步初始化idx=left
+	// right code
+	// int i, idx = left, maxVal = nums[left]; // 需同步初始化idx=left
+    // for (i = left + 1; i <= right; i++) {   // 否则会报内存越界 
+
+    int i, idx, maxVal = INT_MIN;
+    for (i = left; i <= right; i++) {
+        if (nums[i] > maxVal) {
+            maxVal = nums[i];
+            idx = i;
+        }
+    }
+    root->val = maxVal;
+
+    // 递归调用生成左右子树
+    root->left = buildMaxTree(nums, left, idx - 1);
+    root->right = buildMaxTree(nums, idx + 1, right);
+
+    return root;
+}
+
+
+struct TreeNode* constructMaximumBinaryTree(int* nums, int numsSize){
+    if (nums == NULL) return NULL;
+
+    return buildMaxTree(nums, 0, numsSize - 1);
+
+}

c_code/704_search.c

@@ -0,0 +1,20 @@
+int search(int* nums, int numsSize, int target){
+    if (nums == NULL) return -1;
+    int left = 0;
+    int right = numsSize - 1; // 左闭右闭区间
+
+    while (left <= right) {
+        int mid = left + (right - left) / 2;
+        
+        if (nums[mid] == target) {
+            return mid;
+        } else if (nums[mid] < target) {
+            left = mid + 1;
+        } else if (nums[mid] > target) {
+            right = mid - 1;
+        }
+    }
+
+    return -1; // 遍历完未找到
+
+}