DFS Construct Binary Tree from Inorder and Postorder Traversal

思路：

DFS，与上一题思考方式一样，这次的root出现在Post-Order的末尾。

Example:

In-Order: 1 2 3 4 5 6

Post-Order: 1 3 2 6 5 4

先从Post-Order末尾找到root，这里是4，然后在In-Order中找到root的位置，左半部分就是root的左子树的元素，右半部分就是root右子树的所有元素，这里1，2，3是root的左子树的元素，5，6是root右子树的元素，依次类推，再构造root的左子树和右子树，并且让root->left = 构造的左子树 & root->right = 构造的右子树。

/**
 * 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 {
private:
    //Construct Binary Tree from Inorder and Postorder Traversal
    template<typename Iter>
    TreeNode* build(Iter iBegin, Iter iEnd, Iter pBegin, Iter pEnd) {
        if(iBegin == iEnd) return NULL;
        if(pBegin == pEnd) return NULL;
        int val = *prev(pEnd);
        auto iRoot = find(iBegin, iEnd, val);
        TreeNode *root = new TreeNode(*iRoot);
        int leftSize = iRoot - iBegin;
        root->left = build(iBegin, iRoot, pBegin, pBegin + leftSize);
        root->right = build(iRoot + 1, iEnd, pBegin + leftSize, prev(pEnd));
        return root;
    }
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        size_t size = postorder.size();
        if(size == 0) return NULL;
        return build(inorder.begin(), inorder.end(), postorder.begin(), postorder.end());
    }
};