[leetCode]Construct Binary Tree from Preorder and Inorder Traversal

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

给了前序和中序遍历，这个题也好做，主要问题是把左右分支的下标给搞清楚。

对于前序来说，本次的根节点总是第一个preorder[0];左子树坐标是[prestart+1,prestart+i],右子树是[prestart+i+1,preend];

而对于中序来说，如果根节点是第i个（从instart开始数），那么左边子树的坐标是[instart,instart+i-1],右子树是[instart+i+1,inend];

如果觉得前序不好算的话，可以先算中序，前序的左子树段是中序左子树段+1（但是要注意初始坐标不一样）

#include <iostream>
#include <vector>
using namespace std;
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) {
        TreeNode * head = build(preorder,0,preorder.size()-1,inorder,0, inorder.size()-1);
        return head;
    }

    TreeNode *build(vector<int> &preorder, int preStart,int preEnd,vector<int> &inorder, int inStart, int inEnd){
        if(preStart == preEnd) return new TreeNode(preorder[preStart]);
        if(preStart > preEnd)    return NULL;
        int head = preorder[preStart];
        int i = 0;
        int size = preEnd - preStart+1;
        for(; i<size;i++){
            if(inorder[i+inStart] == head){
                break;
            }
        }
        TreeNode *node = new TreeNode(head);
        node->left = build(preorder,preStart+1,preStart+i,inorder, inStart, inStart+i-1);
        node->right = build(preorder, preStart+1+i,preEnd,inorder,inStart+1+i,inEnd);
        return node;

    }
};