"Coding Interview Guide" -- 先序、中序和后序数组两两结合重构二叉树

　　import java.util.HashMap;

 1 public Node preInToTree(int[] pre, int[] in)
 2 {
 3     if(pre == null || in == null)
 4     {
 5         return null;
 6     }
 7 
 8     HashMap<Integer, Integer> map = new HashMap<>();        // 哈希表的键值对用于记录元素及其索引
 9     for(int i = 0; i < in.length; i++)
10     {
11         map.put(in[i], i);
12     }
13         
14     return preIn(pre, 0, pre.length - 1, in, 0, in.length - 1, map);
15 }
16 
17 public Node preIn(int[] p, int pi, int pj, int[] n, int ni, int nj, HashMap<Integer, Integer> map)
18 {
19     if(pi > pj)
20     {
21         return null;
22     }
23 
24     Node head = new Node(p[pi]);            // 当前先序数组的第一个元素是当前子树的头节点
25     int index = map.get(p[pi]);
26     head.left = preIn(p, pi + 1, pi + index - ni, n, ni, index - 1, map);
27     head.right = preIn(p, pi + index - ni + 1, pj, n, index + 1, nj, map);
28     return head;
29 }

 1 public Node inPosToTree(int[] in, int[] pos)
 2 {
 3     if(in == null || pos == null)
 4     {
 5         return null;
 6     }
 7 
 8     HashMap<Integer, Integer> map = new HashMap<>();             // 哈希表的键值对用于记录元素及其索引
 9     for(int i = 0; i < in.length; i++)
10     {
11         map.put(in[i], i);
12     }
13 
14     return inPos(in, 0, in.length - 1, pos, 0, pos.length - 1, map);
15 }
16 
17 public Node inPos(int[] n, int ni, int nj, int[] s, int si, int sj, HashMap<Integer, Integer> map)
18 {
19     if(si > sj)
20     {
21         return null;
22     }
23 
24     Node head = new Node(s[sj]);           // 当前后序数组的最后一个元素是当前子树的头节点
25     int index = map.get(s[sj]);
26     head.left = inPos(n, ni, index - 1, s, si, si + index - ni - 1, map);
27     head.right = inPos(n, index + 1, nj, s, si + index - ni, sj - 1, map);
28     return head;
29 }