【题目描述】Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

【解题思路】与上题相似

【考查内容】树，递归

/**
 * 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 *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if(preorder.size()!=inorder.size()) 
            return NULL;
        return build(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
    }
    TreeNode* build(vector<int> &preorder,int l1,int r1,vector<int> &inorder,int l2,int r2){
        if(l1>r1||l2>r2) 
            return NULL;
        int r = preorder[l1],i,cnt=0;
        for(i=l2;inorder[i]!=r;i++,cnt++);
        TreeNode * root = new TreeNode(preorder[l1]);
        root->left = build(preorder,l1+1,l1+cnt,inorder,l2,i-1);
        root->right = build(preorder,l1+cnt+1,r1,inorder,i+1,r2);
        return root;
    }
};
`