106. 从中序与后序遍历序列构造二叉树

 

 思路：和105题一样，这次根节点在后续遍历的最后，找到它；之后在中序遍历里找到根节点。Arrays.copyOfRange()函数找出中序数组和后续数组的左右子树序列，递归，构成树，返回根节点。。。不细说了。感觉挺简单。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder.length>0&&postorder.length>0)
     {
        TreeNode root=new TreeNode(postorder[postorder.length-1]);
        int num=0;//根节点在中序遍历里的下标
        for(int i=0;i<inorder.length;i++)
        {
            if(inorder[i]==root.val)
                {
                    num=i;//左子树加上根节点有num+1个，左子树上面num个元素，这里注意
                    break;
                }
        }
           int []inLeft=Arrays.copyOfRange(inorder,0,num);
            int[] inRight=Arrays.copyOfRange(inorder,num+1,inorder.length);
            int []postLeft=Arrays.copyOfRange(postorder,0,num);
            int []postRight=Arrays.copyOfRange(postorder,num,postorder.length-1);
            root.left=buildTree(inLeft,postLeft);
            root.right=buildTree(inRight,postRight);
            return root;
     }
       else
        {
            return null;
        }
    }
}