[LeetCode]Construct Binary Tree from Inorder and Postorder Traversal

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.

思路：  中序的是 左中右，后序的是 左右中 ，  后序的最后一个元素即是根， 之后去中序中查找根的位置，即能求出 左右子树的 序列长度，

进而递归解子问题。

下面的代码存在两个问题，一是空间复杂度高，大数据内存超了，二是如果后序的根元素在中序中多次出现，可能会查找错误，那么要把中序的左子树和后序的左子树进行比较，若排序之后一致，才能采用，本题似乎没有出现这样的问题。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *help(vector<int> &inorder, vector<int> &postorder){
        if(postorder.size()==0)
			return NULL;
		if(postorder.size()==1)
		{
			TreeNode *p = new TreeNode(inorder[0]);
			return p;
		}
		int len = postorder.size();
		int root = postorder[len-1];
//		cout << root << endl;
		TreeNode *p = new TreeNode( root);
        int site;
         for(int i = 0 ; i < inorder.size()  ; i++)
	         if (inorder[i]==root)
             {
                 site = i;
                 break;
             }
		vector<int> inorder_l,inorder_r;
		vector<int> postorder_l,postorder_r;
		postorder_l.clear();postorder_r.clear();
		inorder_l.clear();inorder_r.clear();
		for(int i = 0 ; i < site ; i++)
		{
			inorder_l.push_back(inorder[i]);
			postorder_l.push_back(postorder[i]);
		}

        if(site+1 != inorder.size())
        {
		for(int i = site ; i+1 < inorder.size()  ; i++)
			postorder_r.push_back(postorder[i]);
		for(int i = site+1;  i < inorder.size() ; i++)
			inorder_r.push_back(inorder[i]);
        }

		p->left = help(inorder_l,postorder_l);
		p->right = help(inorder_r,postorder_r);
		return p;
		
	}
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
       TreeNode *p = help(inorder,postorder);
	   return p;
    }
};

传下标，节约空间

class Solution {
public:
    TreeNode *help(vector<int> &inorder, int a, int b,vector<int> &postorder,int c , int d){
    	
		int len = b-a+1;
		if(len==0)
			return NULL;
		if(len==1)
		{
			TreeNode *p = new TreeNode(inorder[a]);
			return p;
		}
		int root = postorder[d];
	
		TreeNode *p = new TreeNode( root);
        int site;
         for(int i = a ; i <= b ; i++)
	         if (inorder[i]==root)
             {
                 site = i;
                 break;
             }
   
		int left_len = site - a;

		
        int right_len = len - left_len -1;
 
        if(left_len>0)
		    p->left = help(inorder,a,site-1,postorder,c, c + left_len-1);
		if(right_len>0)
            p->right = help(inorder,site+1, b,postorder,c+left_len, d-1);
		return p;
		
	}
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
		size_t len = inorder.size();
		if(len == 0)
			return NULL;
       TreeNode *p = help(inorder,0,len-1,postorder,0,len-1);
	   return p;
    }
};