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

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

 

和上一道题基本一样

 

根据中序和后序遍历确定一个棵树。

 

/**
 * 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) {
      

        int len = inorder.length;
        if( len == 0)
            return null;

        return helper(inorder,0,len-1,postorder,len-1);

    }

    public TreeNode helper(int[] inorder,int start,int end, int[] postorder, int pos ){

        if( pos < 0 || start > end)
            return null;
        TreeNode node = new TreeNode(postorder[pos]);

        int size = 0;
        for( int i = end;i >= start && inorder[i] != postorder[pos];i--,size++)
            ;

        node.left = helper(inorder,start,end-size-1,postorder,pos-size-1);

        node.right = helper(inorder,end-size+1,end,postorder,pos-1);

        return node;

    }
}