LeetCode: Construct Binary Tree from Preorder and Inorder Traversa

跟inorder and postorder基本一样，一次过吧

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