Construct Binary Tree from Preorder and Inorder Traversal——LeetCode

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

题目大意：给定一个二叉树的前序和中序序列，构建出这个二叉树。

解题思路：跟中序和后序一样，在先序中取出根节点，然后在中序中找到根节点，划分左右子树，递归构建这棵树，退出条件就是左右子树长度为1，则返回这个节点，长度为0则返回null。

Talk is cheap：

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || inorder == null || inorder.length == 0 || preorder.length == 0) {
            return null;
        }
        int preLen = preorder.length;
        int inLen = inorder.length;
        TreeNode root = new TreeNode(preorder[0]);
        if (preLen == 1) {
            return root;
        }
        int pos = 0;
        for (int i = 0; i < inLen; i++) {
            if (inorder[i] == preorder[0]) {
                pos = i;
                break;
            }
        }

        int[] preLeft = Arrays.copyOfRange(preorder, 1, pos + 1);
        int[] inLeft = Arrays.copyOfRange(inorder, 0, pos);
        int[] preRight = Arrays.copyOfRange(preorder, pos + 1, preLen);
        int[] inRight = Arrays.copyOfRange(inorder, pos + 1, inLen);
        root.left = buildTree(preLeft, inLeft);
        root.right = buildTree(preRight, inRight);
        return root;
    }