#Problem description:

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

#Solution:
This problem is pretty similar to 105. Construct Binary Tree from Preorder and Inorder Traversal.

We can see the inorder and postorder array like this:

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           in order
+------------+------+-------------+
| left child | root | right child |
+------------+------+-------------+  

            post order
+------------+-------------+------+
| left child | right child | root |
+------------+-------------+------+

As you can see, root is in the last position in postorder array. After we find the root, we can get the length of each subarray as follows:

  • left tree:
    inOrder[1 .. p - 1]
    postOrder[1 .. p - 1]
  • right tree:
    inOrder[p + 1 .. n]
    postOrder[p .. n - 1]
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/**
 * 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 deal(0, inorder.size() - 1, 0, postorder.size() - 1, inorder, postorder);
    }
    
    TreeNode* deal(int leftInorder, int rightInorder, int leftPostorder, int rightPostorder, vector<int>& inorder, vector<int>& postorder) {
        if (rightInorder < leftInorder || rightPostorder < leftPostorder) return NULL;
        
        TreeNode* rt = new TreeNode(postorder[ rightPostorder ]);

        int p = 0;
        for (int i = 0; i <= rightInorder - leftInorder; ++i)
            if (inorder[leftInorder + i] == rt -> val) {
                p = i; break;
            }

        rt -> left  = deal(leftInorder, leftInorder + p - 1, leftPostorder, leftPostorder + p - 1, inorder, postorder);
        rt -> right = deal(leftInorder + p + 1, rightInorder, leftPostorder + p, rightPostorder - 1, inorder, postorder);

        return rt;
    }
    
};