106. 從中序與後序周遊序列構造二叉樹
難度中等462
根據一棵樹的中序周遊與後序周遊構造二叉樹。
注意:
你可以假設樹中沒有重複的元素。
例如,給出
傳回如下的二叉樹:中序周遊 inorder = [9,3,15,20,7] 後序周遊 postorder = [9,15,7,20,3]
3 / \ 9 20 / \ 15 7
和前序周遊類似,将後序周遊到着過來,從後往前
/**
* 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 post_size;
unordered_map<int, int> index;
TreeNode* myBuildTree(const vector<int>& inorder, const vector<int>& postorder, int in_left, int in_right){
if(in_left > in_right){
return nullptr;
}
int root_val = postorder[post_size];
TreeNode* root = new TreeNode(root_val);
int id = index[root_val];
post_size--;
root->right = myBuildTree(inorder, postorder, id+1, in_right);
root->left = myBuildTree(inorder, postorder, in_left, id-1);
return root;
}
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
post_size = postorder.size()-1;
for(int i = 0; i < inorder.size(); i++){
index[inorder[i]] = i;
}
return myBuildTree(inorder, postorder, 0, inorder.size()-1);
}
};