Leetcode dfs Construct Binary Tree from Inorder and Postorder Traversal

Construct Binary Tree from Inorder and Postorder Traversal

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

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

题意：从前序遍历和中序遍历结果重构二叉树
思路1：递归
从后序遍历结果可以得到根节点
从根节点和中序遍历结果可以得到左右子树
按上面两个规律递归就可以重构二叉树了

复杂度：时间O(n) --> 总共有n个节点要构建，空间O(log n) --> 最多需要保存前面 log n个状态，每个状态需要保存的变量是O(1)
思路2：非递归 --> todo
思路2和思路1一样，不过采用非递归的方法实现
复杂度：

vector<int> _postorder;
vector<int> _inorder;
void dfs(int post_start, int post_end, int in_start, int in_end, TreeNode *&root){
	if(post_start >= post_end)return;
	root = new TreeNode(_postorder[post_end - 1]);
	int i;
	for(i = in_start; i < in_end; ++i){
		if(_inorder[i] == _postorder[post_end - 1]) break;
	}
	dfs(post_start, post_start + i - in_start, in_start, in_start + i, root->left);
	dfs(post_start + i - in_start, post_end - 1, i + 1, in_end, root->right);
}
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
	TreeNode *root = NULL;
	_postorder = postorder, _inorder = inorder;
	dfs(0, postorder.size(), 0, inorder.size(), root);
	return root;
}