105. 从前序与中序遍历序列构造二叉树 - 力扣（LeetCode）
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int n;
unordered_map<int, int> pos;
TreeNode* build(vector<int> &preorder, vector<int> &inorder, int il, int ir, int pl, int pr)
{
TreeNode* root = new TreeNode(preorder[pl]);
int k = pos[root->val];
if (il < k) //如果左子树存在，递归构建左子树
root->left = build(preorder, inorder, il, k - 1, pl + 1, pl + 1 + k - 1 - il);
if (ir > k) //如果右子树存在，递归构建右子树
root->right = build(preorder, inorder, k + 1, ir, pl + 1 + k - 1 - il + 1, pr);
return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
n = preorder.size();
for (int i = 0; i < n; i++) pos[inorder[i]] = i;
return build(preorder, inorder, 0, n - 1, 0, n - 1);
}
};
106. 从中序与后序遍历序列构造二叉树 - 力扣（LeetCode）
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int n;
unordered_map<int, int> pos;
TreeNode* build(vector<int> &inorder, vector<int> &postorder, int il, int ir, int pl, int pr)
{
TreeNode* root = new TreeNode(postorder[pr]);
int k = pos[root->val];
if (il < k)
root->left = build(inorder, postorder, il, k - 1, pl, pl + k - 1 - il);
if (ir > k)
root->right = build(inorder, postorder, k + 1, ir, pl + k - 1 - il + 1, pr - 1);
return root;
}
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
n = inorder.size();
for (int i = 0; i < n; i++) pos[inorder[i]] = i;
return build(inorder, postorder, 0, n - 1, 0, n - 1);
}
};