[LeetCode] 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.

 

方法：在inorder中寻找postorder的最后一个，然后左右递归，注意处理好各个low，high边界

class Solution
{
    public:
        TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder)
        {
            return buildTree(postorder, 0, postorder.size()-1,
                              inorder, 0, inorder.size()-1);
        }

        TreeNode *buildTree(vector<int> &postorder, int low1, int high1,
                            vector<int> &inorder, int low2, int high2)
        {
            //cout << "==============" <<endl;
            //cout << "low1 = \t" << low1 <<endl;
            //cout << "high1= \t" << high1 <<endl;
            //cout << "low2 = \t" << low2 <<endl;
            //cout << "high2= \t" << high2 <<endl;
            if(low1 > high1 || low2> high2)
                return NULL;

            TreeNode * p = new TreeNode(postorder[high1]);
            if(low1 == high1)
            {   
                return p;
            }   
            int index = 0;
            for(index = low2; index < high2; index++)
            {   
                if(inorder[index] == postorder[high1])
                    break;
            }   
            //cout << "index= \t" << index<<endl;

            if(index != low2)
                p->left = buildTree(postorder, low1,(low1) + (index-1-low2), inorder, low2, index-1);
            if(index != high2)
                p->right = buildTree(postorder, (high1-1) - (high2-index-1) ,high1-1, inorder, index+1, high2);

            return p;
        }
} ;