From 5cb33c319b8652f2694fccc9db88127ce5f24563 Mon Sep 17 00:00:00 2001
From: Arthur <gpan20020205@gmail.com>
Date: Wed, 20 Oct 2021 16:10:04 +0100
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---
 ...序与后序遍历序列构造二叉树.md | 75 +++++++++++++++++++
 1 file changed, 75 insertions(+)

diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md
index 39b55d8c..9e0b2430 100644
--- a/problems/0106.从中序与后序遍历序列构造二叉树.md
+++ b/problems/0106.从中序与后序遍历序列构造二叉树.md
@@ -818,6 +818,81 @@ var buildTree = function(preorder, inorder) {
 };
 ```
 
+## C
+106 从中序与后序遍历序列构造二叉树
+```c
+int linearSearch(int* arr, int arrSize, int key) {
+    int i;
+    for(i = 0; i < arrSize; i++) {
+        if(arr[i] == key)
+            return i;
+    }
+    return -1;
+}
+
+struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize){
+    //若中序遍历数组中没有元素，则返回NULL
+    if(!inorderSize)
+        return NULL;
+    //创建一个新的结点，将node的val设置为后序遍历的最后一个元素
+    struct TreeNode* node = (struct TreeNode*)malloc(sizeof(struct TreeNode));
+    node->val = postorder[postorderSize - 1];
+
+    //通过线性查找找到中间结点在中序数组中的位置
+    int index = linearSearch(inorder, inorderSize, postorder[postorderSize - 1]);
+
+    //左子树数组大小为index
+    //右子树的数组大小为数组大小减index减1（减的1为中间结点）
+    int rightSize = inorderSize - index - 1;
+    node->left = buildTree(inorder, index, postorder, index);
+    node->right = buildTree(inorder + index + 1, rightSize, postorder + index, rightSize);
+    return node;
+}
+```
+
+105 从前序与中序遍历序列构造二叉树
+```c
+struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize){
+    // 递归结束条件：传入的数组大小为0
+    if(!preorderSize)
+        return NULL;
+
+    // 1.找到前序遍历数组的第一个元素， 创建结点。左右孩子设置为NULL。
+    int rootValue = preorder[0];
+    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
+    root->val = rootValue;
+    root->left = NULL;
+    root->right = NULL;
+
+    // 2.若前序遍历数组的大小为1，返回该结点
+    if(preorderSize == 1)
+        return root;
+
+    // 3.根据该结点切割中序遍历数组，将中序遍历数组分割成左右两个数组。算出他们的各自大小
+    int index;
+    for(index = 0; index < inorderSize; index++) {
+        if(inorder[index] == rootValue)
+            break;
+    }
+    int leftNum = index;
+    int rightNum = inorderSize - index - 1;
+
+    int* leftInorder = inorder;
+    int* rightInorder = inorder + leftNum + 1;
+
+    // 4.根据中序遍历数组左右数组的各子大小切割前序遍历数组。也分为左右数组
+    int* leftPreorder = preorder+1;
+    int* rightPreorder = preorder + 1 + leftNum; 
+
+    // 5.递归进入左右数组，将返回的结果作为根结点的左右孩子
+    root->left = buildTree(leftPreorder, leftNum, leftInorder, leftNum);
+    root->right = buildTree(rightPreorder, rightNum, rightInorder, rightNum);
+
+    // 6.返回根节点
+    return root;
+}
+```
+
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
 * B站视频：[代码随想录](https://space.bilibili.com/525438321)