Construct Binary Tree from Inorder and Preorder（Inorder and Postorder） Traversal

题目：

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

Note:
You may assume that duplicates do not exist in the tree.

思想：

由树的遍历可知，由中序遍历与后序遍历，或是由中序遍历与先序遍历，均能够确定一棵二叉树。以先序遍历为例，首先根据先序遍历找到根节点，即先序遍历的preorder[0]，然后在中序遍历中找到与preorder[0]相等的数inorder[i] 。这样就把inorder分成了两部分，inorder[0] ~inorder[i-1] ,构成preorder [0] 的左子树，inorder[i+1]~inorder[end]构成preorder[0]的右子树。preorder也被分成两部分，preorder[1]~preorder[1+i] 和preorder[i+2]~preorder[end]两部分，分别为左子树和右子树两个部分。然后递归的构建左子树和右子树即可。后序遍历只要以postorder[end]为根即可。

代码：

/*** 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* build(vector<int> &inorder, int istart, int iend, vector<int> &postorder, int pstart, int pend){TreeNode *T = new TreeNode(postorder[pend]);if (pstart == pend && istart == iend){return T;}int k = 0;for (int i = istart; i <= iend; i++){if (inorder[i] == postorder[pend])break;elsek++;}if ((k - 1) >= 0)T->left = build(inorder, istart, istart + k - 1, postorder, pstart, pstart + k - 1);if ((iend - istart - k - 1) >= 0)T->right = build(inorder, istart + k + 1, iend, postorder, pstart + k , pend-1);return T;}TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) //后序和中序{TreeNode *T = NULL;if (inorder.size() == NULL || postorder.size() == NULL){return T;}T = build(inorder, 0,inorder.size() - 1, postorder, 0, postorder.size() - 1);return T;}void preorder(TreeNode* root){if (root != NULL){cout << root->val << " ";preorder(root->left);preorder(root->right);}}};TreeNode* build(vector<int> &preorder, int pstart, int pend, vector<int> &inorder, int istart, int iend){TreeNode *T = new TreeNode(preorder[pstart]);if (pstart==pend && istart==iend){return T ;}int k = 0;for (int  i = istart; i <= iend; i++){if (inorder[i] == preorder[pstart])break;elsek++;}if ((k-1)>=0)T->left = build(preorder, pstart + 1, pstart + k, inorder, istart, istart + k - 1);if ( (pend - pstart - k -1) >= 0)T->right = build(preorder, pstart + k + 1, pend, inorder, istart + k + 1, iend);return T;}TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) //先序和中序{TreeNode *T = NULL;if (preorder.size() == NULL || inorder.size() == NULL){return T;}T = build(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);return T;}