Leetcode106. Construct Binary Tree from Inorder and Postorder Traversal中序后续构造二叉树

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

 

 

class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) 
    {
        if(postorder.size() == 0)
            return NULL;
        if(postorder.size() == 1)
            return new TreeNode(postorder[0]);
        int iroot = postorder[postorder.size() - 1];
        int ipos = 0;
        for(int i = 0; i < inorder.size(); i++)
        {
            if(inorder[i] == iroot)
            {
                ipos = i;
                break;
            }
        }
        TreeNode *root = new TreeNode(iroot);
        vector<int> v1(postorder.begin(), postorder.begin() + ipos);
        vector<int> v2(inorder.begin(), inorder.begin() + ipos);
        vector<int> v3(postorder.begin() + ipos, postorder.end() - 1);
        vector<int> v4(inorder.begin() + ipos + 1, inorder.end());
        root ->left = buildTree(v2, v1);
        root ->right = buildTree(v4, v3);
        return root;
    }
};