给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

思路一：递归

struct TreeNode* createTreeNode(int val) {struct TreeNode* ret = malloc(sizeof(struct TreeNode));ret->val = val;ret->left = ret->right = NULL;return ret;
}struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize) {if (postorderSize == 0) {return NULL;}struct TreeNode* root = createTreeNode(postorder[postorderSize - 1]);struct TreeNode** s = malloc(sizeof(struct TreeNode*) * 10001);int top = 0;s[top++] = root;int inorderIndex = inorderSize - 1;for (int i = postorderSize - 2; i >= 0; i--) {int postorderVal = postorder[i];struct TreeNode* node = s[top - 1];if (node->val != inorder[inorderIndex]) {node->right = createTreeNode(postorderVal);s[top++] = node->right;} else {while (top > 0 && s[top - 1]->val == inorder[inorderIndex]) {node = s[--top];inorderIndex--;}node->left = createTreeNode(postorderVal);s[top++] = node->left;}}return root;
}

分析：

本题要利用二叉树的中序遍历和后序遍历来确定二叉树，即可不断创建新二叉树将后序遍历的右子树赋值给新二叉树，不断创建，等栈顶为根节点的位置时再将左子树创建为新二叉树最后输出

总结:

本题考察对二叉树的应用，先找到根节点，不断添加二叉树即可解决