105. Construct Binary Tree from Preorder and Inorder Traversal

题目

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

思路

1. 前序遍历的第一个为root节点（3），找到中序遍历该节点，以此划分左右子树； 9|3|15，20，7；
2. 根据上面的划分同时划分前序数组 3|9|20,15，7，则判断出左子树是一个单独的9，而右子树的根节点为20；
3. 根据中序遍历的特点继续第一步，根据右子树的根节点（20）划分右子树的左右-->递归
   递归：左子树sub数组，右子树sub数组

代码

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        TreeNode root = null;
        if (preorder.length == 0 || inorder.length == 0) {
            return root;
        }
        root = new TreeNode(preorder[0]);
        int tmp = 0;
        for (int i = 0; i < inorder.length; i++) {
            tmp = i;
            if(preorder[0] == inorder[i]){
                break;
            }
        }
        int[] leftInorder = Arrays.copyOfRange(inorder,0,tmp);
        int[] rightInorder = Arrays.copyOfRange(inorder,tmp+1,inorder.length);
        int[] leftPreorder = Arrays.copyOfRange(preorder,1,1+tmp);
        int[] rightPreorder = Arrays.copyOfRange(preorder,tmp+1,preorder.length);
        root.left = buildTree(leftPreorder,leftInorder);
        root.right = buildTree(rightPreorder,rightInorder);
        return root;
    }