[LeetCode]106. Construct Binary Tree from Inorder and Postorder Traversal

106. Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

/**
* 和之前使用中序和前序遍历重建的题一样，可以直接使用索引，避免反复创建vector，这里就省略不实现了...
*/
/**
* 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:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
if(inorder.empty() || postorder.empty())
return NULL;
TreeNode* root = new TreeNode(*(postorder.end()-1));
int i = 0;
for(; i < inorder.size(); ++i){
if(inorder[i] == root->val)
break;
}
if(i > 0){
vector<int> in_left = vector<int>(inorder.begin(), inorder.begin()+i);
vector<int> post_left = vector<int>(postorder.begin(), postorder.begin()+i);
root->left = buildTree(in_left, post_left);
}
if(i < inorder.size()-1){
vector<int> in_right = vector<int>(inorder.begin()+i+1, inorder.end());
vector<int> post_right = vector<int>(postorder.begin()+i, postorder.end()-1);
root->right = buildTree(in_right, post_right);
}
return root;
}
};