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

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

一刷
思路同105

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
       return helper(inorder, 0, inorder.length-1, postorder, inorder.length-1);
    }
    
    private TreeNode helper(int[] inorder, int inStart, int inEnd, int[] postorder, int pEnd){
        if(inStart>inEnd) return null;
        if(inEnd>=inorder.length) return null;
        if(pEnd >= postorder.length) return null;
        
        TreeNode root = new TreeNode(postorder[pEnd]);
        int index = 0;
        for(index = inStart; index<= inEnd; index++){
            if(inorder[index] == root.val) break;
        }
        root.left = helper(inorder, inStart, index-1, postorder, pEnd-inEnd+index-1);
        root.right = helper(inorder, index+1, inEnd, postorder, pEnd-1);
        return root;
    }
}