AcWing 18. 重建二叉树
原题链接
中等
作者:
小.bug
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2022-08-10 17:05:19
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所有人可见
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阅读 314
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> preorder, inorder;
unordered_map<int, int> pos;
TreeNode* build(int pl, int pr, int il, int ir)
{
if(pl > pr) return NULL;
TreeNode* root = new TreeNode(preorder[pl]);
int k = pos[root->val];
root->left = build(pl + 1, pl + 1 + k - 1 - il, il, k - 1);
root->right = build(pl + 1 + k - 1 - il + 1, pr, k + 1, ir);
return root;
}
TreeNode* buildTree(vector<int>& _preorder, vector<int>& _inorder) {
preorder = _preorder, inorder = _inorder;
int n = _inorder.size();
for(int i = 0; i < n; i ++) pos[inorder[i]] = i;
TreeNode* root = build(0, n - 1, 0, n - 1);
return root;
}
};