construct-binary-tree-from-preorder-and-inorder-traversal

1、来源：construct-binary-tree-from-preorder-and-inorder-traversal
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

2、思路：

• 先序的第一个元素是根结点，在中序中，该结点将数组分为左右两部分，分别代表左右子树，再递归即可；

3、代码：

  public TreeNode buildTree(int[] preorder, int[] inorder) {        if(preorder == null || preorder.length == 0){            return null;        }        return biuld(preorder, 0, preorder.length - 1,inorder, 0, inorder.length - 1);    }    private TreeNode biuld(int[] preorder, int start1, int end1, int[] inorder, int start2,            int end2) {        int root_val = preorder[start1];        TreeNode root = new TreeNode(root_val);        int mid;        for(mid = start2; mid < end2; mid ++){            if(inorder[mid] == root_val)                break;        }        int left_num = mid - start2;//左子树节点数目        int right_num = end2 - mid;//右子树节点数目        if(left_num > 0){            root.left = biuld(preorder, start1 + 1, start1 + left_num, inorder, start2, mid - 1);        }        if(right_num > 0){            root.right = biuld(preorder, start1 + left_num + 1, end1, inorder, mid + 1, end2);        }        return root;    }