106. Construct Binary Tree from Inorder and Postorder Traversal

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对于后序遍历来说，最后一个元素一定是根节点。然后在中序遍历中找的其未知，左边是左子树，右边是右子树。

左子树在后序遍历中的最后一个节点又是根节点，从而递归下去。

/** * 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) {        if(postorder.empty())            return NULL;                for(int i = 0; i < inorder.size();i++)        {            m[inorder[i]] = i;        }                reutrn build(inorder,0,inorder.size() - 1,postorder,0,postorder.size() - 1);            }        TreeNode* build(vector<int>& inorder,int s0,vector<int>& postorder,int s1,int e1)    {        if(s0 > e0 || s1 > e1)        {            return NULL;        }                TreeNode* root = new TreeNode(postorder[e1]);                int mid = m[postorder[e1]];        int num = mid - s0;                root->left = build(inorder,s0,mid - 1,postorder,s1,s1 + num - 1);        root->right = build(inorder,s0,mid + 1,postorder,s1 + num + 1,e1 - 1);            }};

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