LeetCode106-从中序与后序遍历序列构造二叉树

根据一棵树的中序遍历与后序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]

返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

一、思路

思路与前序遍历很相似：LeetCode105-从前序与中序遍历序列构造二叉树

C++代码：

class Solution {
public:
	TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
		return generateTree(0, inorder.size(), 0, postorder.size(), inorder, postorder);
	}

	TreeNode* generateTree(int in_begin, int in_end, int post_begin, int post_end, vector<int>& inorder, vector<int>& postorder) {
		if (in_begin >= in_end || post_begin >= post_end)
			return NULL;
		int left_begin_in, left_end_in, left_begin_post, left_end_post;
		int right_begin_in, right_end_in, right_begin_post, right_end_post;
		int root_in = in_begin, root_post = post_end - 1;
		TreeNode* root = new TreeNode(postorder[root_post]);

		while (root_in < in_end && inorder[root_in] != postorder[root_post])
			root_in++;
		
		left_begin_in = in_begin;
		left_end_in = root_in;

		right_begin_in = root_in + 1;
		right_end_in = in_end;

		int left = left_end_in - left_begin_in, right = right_end_in - right_begin_in;
		
		right_end_post = root_post;
		right_begin_post = right_end_post - right;

		left_begin_post = post_begin;
		left_end_post = left_begin_post + left;

		root->left = generateTree(left_begin_in, left_end_in, left_begin_post, left_end_post, inorder, postorder);
		root->right = generateTree(right_begin_in, right_end_in, right_begin_post, right_end_post, inorder, postorder);

		return root;
	}
};

执行效率：