Given the levelorder and inorder traversal sequence of a binary tree, reconstruct the original tree.

Assumptions

• The given sequences are not null and they have the same length
• There are no duplicate keys in the binary tree

Examples

levelorder traversal = {5, 3, 8, 1, 4, 11}

inorder traversal = {1, 3, 4, 5, 8, 11}

the corresponding binary tree is

    5

  /    \

3        8

/ \

1 4 11

Solution: level[]第一个为root, 找到该数值在in[]的位置（利用hashmap），再把level[]剩下的拆分为左半部和右半部两个list(根据index大小)

  public TreeNode reconstruct(int[] in, int[] level) {
    if (in.length == 0) {
      return null;
    }
    Map<Integer, Integer> hashmap = new HashMap<>();
    List<Integer> levell = new ArrayList<>();
    for (int i = 0; i < in.length; i++) {
      hashmap.put(in[i], i);
      levell.add(level[i]);
    }
    return build(in, 0, in.length - 1, levell, hashmap);
  }
  private TreeNode build(int[] in, int instart, int inend, List<Integer> level, Map<Integer, Integer> hashmap) {
    if (instart > inend) {
      return null;
    }
    TreeNode root = new TreeNode(level.get(0));
    // int position = find(inorder, instart, inend, post[postend]);
    List<Integer> left = new ArrayList<>();
    List<Integer> right = new ArrayList<>();
    int position = hashmap.get(level.get(0));
    for (int i = 1; i < level.size(); i++) {
      int num = level.get(i);
      int pos = hashmap.get(num);
      if (pos < position) {
        left.add(num);
      } else {
        right.add(num);
      }
    }
    root.left = build(in, instart, position - 1, left, hashmap);
    root.right = build(in, position + 1, inend, right, hashmap);
    return root;
  }