leetcode 105题，由树的前序序列和中序序列构建树结构。详细解答参考《剑指offer》page56.
先序遍历结果的第一个节点为根节点，在中序遍历结果中找到根节点的位置。然后就可以将问题拆分，递归求解。
/**
* 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 * buildTreeCore(vector<int>& preorder, vector<int>& inorder, int p1, int p2, int p3, int p4)
{
int rootValue = preorder[p1];
TreeNode * root = new TreeNode(rootValue);
if (p1 == p2)
{
if (p3 == p4 && preorder[p1] == inorder[p3])
{
return root;
} }
int pr = p3;
while (pr <= p4 && inorder[pr] != rootValue)
pr++; int leftLength = pr - p3;
int p5 = p1 + leftLength;
if (leftLength > )
{
root->left = buildTreeCore(preorder, inorder, p1 + , p5, p3, pr - );
}
if (leftLength < p2 - p1)
{
root->right = buildTreeCore(preorder, inorder, p5 + , p2, pr + , p4);
} return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if (preorder.size() <= || inorder.size() <= )
{
return NULL;
}
return buildTreeCore(preorder, inorder, , preorder.size()-, , inorder.size()-);
}
};