【construct-binary-tree-from-preorder-and-inorder-traversal】

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

Note: 

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

题意：中序和后序遍历重构二叉树

class Solution {public:TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder){int inLen = inorder.size();int postLen = postorder.size();return dfs(inorder, 0, inLen-1, postorder, 0, postLen-1);}TreeNode* dfs(vector<int>& inorder, int inStart, int inEnd,vector<int>& postorder, int postStart, int postEnd){if (inStart>inEnd){return NULL;}TreeNode* root = new TreeNode(postorder[postEnd]);int mid;for (mid=inStart; mid<inEnd; mid++){if (inorder[mid]==root->val){break;}}int leftLen = mid - inStart;root->left = dfs(inorder, inStart, mid - 1, postorder, postStart, postStart + leftLen - 1);root->right = dfs(inorder, mid + 1, inEnd, postorder, postStart + leftLen, postEnd - 1);}};