106 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.

Solution

/**
 * 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* help(vector<int>& inorder, int is, int ie, vector<int>& postorder, int ps, int pe) {
        if ( is > ie ) return NULL;
        int i;
        for ( i = is; i <= ie-1; i++ ) if ( inorder[i] == postorder[pe] ) break;
        TreeNode* ptree = new TreeNode(postorder[pe]);
        ptree->left = help(inorder, is, i-1, postorder, ps, ps-is+i-1);
        ptree->right = help(inorder, i+1, ie, postorder, ps-is+i, pe-1);
        return ptree;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return help(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);
    }
};

Notes
Use sub-vector constructor may encounter "exceed memory limit" problem.