Day-27-Construct Binary Tree from Inorder and Postorder Traversal.cpp

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

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

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7






/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode *Tree(vector<int>& in, int x, int y,vector<int>& po,int a,int b){
        if(x > y || a > b)return nullptr;
        TreeNode *node = new TreeNode(po[b]);
        int SI = x;
        while(node->val != in[SI])SI++;
        node->left  = Tree(in,x,SI-1,po,a,a+SI-x-1);
        node->right = Tree(in,SI+1,y,po,a+SI-x,b-1);
        return node;
    }
    TreeNode* buildTree(vector<int>& in, vector<int>& po){
        return Tree(in,0,in.size()-1,po,0,po.size()-1);
    }
};