Leetcode dfs Construct Binary Tree from Preorder and Inorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal

Total Accepted: 14824 Total
Submissions: 55882My Submissions

Given preorder and inorder 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

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