Construct Binary Tree from Inorder and Postorder Traversal

Construct Binary Tree from Inorder and Postorder 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.

解析：

找到中序和后序遍历规律，后序的最后一个元素是根节点。。。

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* buildTree(vector<int>& inorder, vector<int>& postorder) {                return build(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1);            }        TreeNode* build(vector<int>& inorder,vector<int>& postorder,int inbegin,int inend,int postbegin,int postend)    {        if (inend<inbegin) return NULL;                TreeNode* root=new TreeNode(postorder[postend]);        if (postbegin==postend)         {            root->left=NULL;            root->right=NULL;            return root;        }        int i=0;        for ( i=inbegin; i<=inend; i++)        {            if (inorder[i]==postorder[postend])            {                break;            }        }                root->left=build(inorder,postorder,inbegin,i-1,postbegin,i-1-inbegin+postbegin);        root->right=build(inorder,postorder,i+1,inend,i-inbegin+postbegin,postend-1);        return root;                    }            };