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.

本题和之前的105题比较类似，解法可以用相同的想法来解决，代码如下：

/**
 * 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(postorder.length-1,0,inorder.length-1,inorder,postorder);
    }
    public TreeNode helper(int endPost,int startIn,int endIn,int[] inorder,int[] postorder){
        if(endPost<0||startIn>endIn) return null;
        TreeNode root = new TreeNode(postorder[endPost]);
        int indexIn = 0;
        for(int i=startIn;i<=endIn;i++){
            if(inorder[i]==root.val){
                indexIn = i;
            }
        }
        root.left  =helper(endPost-1-(endIn-indexIn),startIn,indexIn-1,inorder,postorder);
        root.right = helper(endPost-1,indexIn+1,endIn,inorder,postorder);
        return root;
    }
}