*Leetcode_construct-binary-tree-from-preorder-and-inorder-traversal

地址：http://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

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

Note:

You may assume that duplicates do not exist in the tree.
思路：可参考上一篇pat的解题报告(后序+中序)，但思路一样，http://blog.csdn.net/flyupliu/article/details/19249315
in_left 和 in_right 都是相对中序vector invec而言的，从invec的in_left到in_right位置创建节点。

当既有左子树又有右子树时，左子树的根节点位置即为当前根节点位置pre_pos加上1. 右子树的根节点位置即为根节点当前位置pre_pos + 左子树长度( i - in_left ) + 1 .

参考代码：

vector<int>prevec, invec;

TreeNode* Construct(int pre_pos, int in_left, int in_right)
{
if(pre_pos>=prevec.size() || in_left>in_right)
{
return NULL;
}
TreeNode* p = new TreeNode(prevec[pre_pos]);
if(in_left==in_right)
return p;
int i = 0;
for(; i < invec.size(); ++i)
{
if(prevec[pre_pos]==invec[i])
break;
}
//left tree null
if(i == in_left)
{
p->right = Construct(pre_pos+1, in_left+1, in_right);
}
//right tree null
else if(i==in_right)
{
p->left = Construct(pre_pos+1, in_left, in_right-1);
}
else
{
p->left = Construct(pre_pos+1, in_left, i-1);
p->right = Construct(pre_pos+i-in_left+1, i+1, in_right);
}
return p;
}

class Solution {
public:
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
if(preorder.empty())
return NULL;
prevec = preorder;
invec = inorder;
return Construct(0, 0, preorder.size()-1);
}
};

//SECOND TRIALclass Solution {private:    TreeNode *construct(int pre[], int in[], int len)    {        if(len<=0)            return NULL;        int root_val = pre[0], offset = 0;        TreeNode* root = new TreeNode(root_val);        while(in[offset] != root_val)            ++offset;        root->left = construct(pre+1, in, offset);        root->right = construct(pre+offset+1, in+offset+1, len-offset-1);        return root;    }public:    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {        int *pre = &preorder[0], *in = &inorder[0];        return construct(pre, in, preorder.size());    }};