106. 从中序与后序遍历序列构造二叉树

根据一棵树的中序遍历与后序遍历构造二叉树。

注意: 你可以假设树中没有重复的元素。

例如，给出

中序遍历 inorder = [9,3,15,20,7] 后序遍历 postorder = [9,15,7,20,3]

返回如下的二叉树：

3 / \ 9 20 / \ 15 7 /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) { return buildTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1); } TreeNode* buildTree(vector<int> &inorder, int iLeft, int iRight, vector<int> &postorder, int pLeft, int pRight) { if (iLeft > iRight || pLeft > pRight) return NULL; TreeNode * cur = new TreeNode(postorder[pRight]); int i = 0; for (i = iLeft; i < inorder.size(); ++i) { if (inorder[i] == cur->val) break; } cur->left = buildTree(inorder, iLeft, i - 1, postorder, pLeft, pLeft + i - iLeft - 1); cur->right = buildTree(inorder, i + 1, iRight, postorder, pLeft + i - iLeft, pRight - 1); return cur; } };