LeetCode 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.

解题思路：这题和105的解题思路类似，主要区别在于：将先序遍历换成后续后，根节点将是每部分孩子节点的最后一个节点，所以只需要改变的只是根节点在遍历数组中的下标。首先得到树的根节点后，在中序数组中区分左右孩子节点，然后计算左右孩子节点的个数，找到左右孩子的根节点在后续遍历数组中的下标。

/** * 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) {        TreeNode root=null;        if(inorder.length==postorder.length){            if(inorder.length>=0)            root= helper(postorder,postorder.length-1,inorder,0,inorder.length);        }        return root;    }    public TreeNode helper(int[] postorder,int index,int [] inorder,int start,int end){                if(index>= postorder.length || index <0 || start >=end){            return null;        }else{            TreeNode root =new TreeNode(postorder[index]);            int median =0;            for(int i=start;i<end;i++){                if(postorder[index]==inorder[i]){                    median = i;                    break;                }            }            root.left=helper(postorder,index-end+median,inorder,start,median);            root.right=helper(postorder,index-1,inorder,median+1,end);            return root;        }    }}