Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

解題思路：最經典的遞歸，已知先序和中序，構造樹

/**
 * 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* build(vector<int>& preorder, vector<int>& inorder, int pleft, int pright, int ileft, int iright){
        if(pleft == pright)return new TreeNode(preorder[pleft]);
        if(pleft > pright)return NULL;
        int pos;
        TreeNode* root = new
 
 TreeNode(preorder[pleft]);
        for(pos = ileft;pos <= iright; pos++){
            if(inorder[pos] == preorder[pleft])break;
        }
        int lefthalf=pos-ileft;
        int righthalf=iright-pos;
        root->left = build(preorder,inorder,pleft+1,pleft+lefthalf,ileft,pos-1);
        root
 
->right = build(preorder,inorder,pright-righthalf+1,pright,pos+1,iright);
        return root;        
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return build(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
    }
};

105. Construct Binary Tree from Preorder and Inorder Traversal