Construct Binary Tree from Preorder and Inorder Traversal

解析： 思路类似根据中序和后序确定二叉树

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return DFS(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
    }
    TreeNode* DFS(vector<int>& preorder, int pleft, int pright, 
                  vector<int>& inorder, int ileft, int iright){
        if (pleft > pright || ileft > iright) return NULL;
        int i = 0;
        for (i = ileft; i <= iright; ++i){ //寻找二叉树的根节点，根据中序遍历划分左右子树
            if (preorder[pleft] == inorder[i])
                break;
        }
        TreeNode* cur = new TreeNode(preorder[pleft]);
        //前序遍历：左子树的节点 pleft+1, pleft+i-ileft
        //中序遍历：左子树的节点 ileft, i-1       
        cur->left = DFS(preorder, pleft+1, pleft+i-ileft, inorder, ileft, i-1);
        //前序遍历：右子树的节点 pleft+i-ileft+1, pright
        //中序遍历：右子树的节点 i+1, iright  
        cur->right = DFS(preorder, pleft+i-ileft+1, pright, inorder, i+1, iright);
        return cur;
    }
};

[1]https://www.cnblogs.com/grandyang/p/4296500.html?spm=a2c4e.10696291.0.0.6f5c19a4UVoRNo