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添加 0106.从中序与后续遍历构造二叉树.md C语言版本
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@ -818,6 +818,81 @@ var buildTree = function(preorder, inorder) {
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};
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```
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## C
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106 从中序与后序遍历序列构造二叉树
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```c
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int linearSearch(int* arr, int arrSize, int key) {
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int i;
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for(i = 0; i < arrSize; i++) {
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if(arr[i] == key)
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return i;
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}
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return -1;
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}
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struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize){
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//若中序遍历数组中没有元素,则返回NULL
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if(!inorderSize)
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return NULL;
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//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
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struct TreeNode* node = (struct TreeNode*)malloc(sizeof(struct TreeNode));
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node->val = postorder[postorderSize - 1];
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//通过线性查找找到中间结点在中序数组中的位置
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int index = linearSearch(inorder, inorderSize, postorder[postorderSize - 1]);
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//左子树数组大小为index
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//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)
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int rightSize = inorderSize - index - 1;
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node->left = buildTree(inorder, index, postorder, index);
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node->right = buildTree(inorder + index + 1, rightSize, postorder + index, rightSize);
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return node;
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}
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```
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105 从前序与中序遍历序列构造二叉树
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```c
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struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize){
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// 递归结束条件:传入的数组大小为0
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if(!preorderSize)
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return NULL;
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// 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。
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int rootValue = preorder[0];
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struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
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root->val = rootValue;
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root->left = NULL;
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root->right = NULL;
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// 2.若前序遍历数组的大小为1,返回该结点
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if(preorderSize == 1)
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return root;
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// 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
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int index;
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for(index = 0; index < inorderSize; index++) {
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if(inorder[index] == rootValue)
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break;
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}
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int leftNum = index;
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int rightNum = inorderSize - index - 1;
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int* leftInorder = inorder;
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int* rightInorder = inorder + leftNum + 1;
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// 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
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int* leftPreorder = preorder+1;
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int* rightPreorder = preorder + 1 + leftNum;
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// 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
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root->left = buildTree(leftPreorder, leftNum, leftInorder, leftNum);
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root->right = buildTree(rightPreorder, rightNum, rightInorder, rightNum);
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// 6.返回根节点
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return root;
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}
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```
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-----------------------
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* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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* B站视频:[代码随想录](https://space.bilibili.com/525438321)
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