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
cpp
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