LeetCode: Construct Binary Tree from Inorder and Postorder Traversal

第一次只用4个系数写dfs怎么写都不对，后来看了网上的答案发现写右子树的时候开始点不一样的。。。所以要用5个系数

/**
 * 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* dfs(vector<int> &inorder, vector<int> &postorder, int inbeg, int postbeg, int len) {
        if (!len) return NULL;
        TreeNode *tmp = new TreeNode(postorder[postbeg+len-1]);
        int leftlen, rightlen;
        for (int i = inbeg; i < inbeg+len; i++) {
            if (tmp->val == inorder[i]) {
                leftlen = i-inbeg;
                rightlen = len - leftlen -1;
            }
        }
        tmp->left = dfs(inorder, postorder, inbeg, postbeg, leftlen);
        tmp->right = dfs(inorder, postorder, inbeg+leftlen+1, postbeg+leftlen, rightlen);
        return tmp;
    }
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (inorder.size() != postorder.size() || inorder.size() == 0) return NULL;
        int len = inorder.size();
        return dfs(inorder, postorder, 0, 0, len);
    }
};

 C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode BuildTree(int[] inorder, int[] postorder) {
        if (inorder.Length != postorder.Length || inorder.Length == 0) return null;
        int len = inorder.Length;
        return dfs(inorder, postorder, 0, 0, ref len);
    }
    TreeNode dfs(int[] inorder, int[] postorder, int inbeg, int postbeg, ref int len)
    {
        if (len == 0) return null;
        TreeNode tmp = new TreeNode(postorder[postbeg+len-1]);
        int leftlen = 0;
        int rightlen = 0;
        for (int i = inbeg; i < inbeg+len; i++)
        {
            if (tmp.val == inorder[i])
            {
                leftlen = i - inbeg;
                rightlen = len - leftlen - 1;
            }
        }
        tmp.left = dfs(inorder, postorder, inbeg, postbeg, ref leftlen);
        tmp.right = dfs(inorder, postorder, inbeg+leftlen+1, postbeg+leftlen, ref rightlen);
        return tmp;
    }
}