不断递归的实现！！！！
下面是AC代码：
1 /**
2 * Given inorder and postorder traversal of a tree, construct the binary tree.
3 * @param inorder
4 * @param postorder
5 * @return
6 */
7 public TreeNode buildTree(int[] inorder,int[] postorder){
8 if(inorder == null || postorder ==null
9 || inorder.length == 0 || postorder.length == 0)
10 return null;
11 if(inorder.length!=postorder.length)
12 return null;
13 if(inorder.length == 1)
14 {
15 if(inorder[0] != postorder[0])
16 return null;
17 else
18 return new TreeNode(inorder[0]);
19 }
20 // make sure how many nodes in the left
21 int i=0;
22 while(inorder[i++] != postorder[postorder.length-1]);
23 int[] leftInorder = new int[i-1];
24 int[] leftPostorder = new int[i-1];
25 int j = 0;
26 while(j<i-1){
27 leftInorder[j] = inorder[j];
28 leftPostorder[j] = postorder[j];
29 j++;
30 }
31 TreeNode left = buildTree(leftInorder, leftPostorder);
32 TreeNode root = new TreeNode(postorder[postorder.length-1]);
33 root.left = left;
34 // find the right subtree nodes
35 int[] rightInorder = new int[inorder.length - (i-1) -1];
36 int[] rightPostorder = new int[postorder.length - (i-1) -1];
37
38 j = i;
39 while(j<inorder.length){
40 rightInorder[j-i] = inorder[j];
41 rightPostorder[j-i] = postorder[j-1];
42 j++;
43 }
44 root.right = buildTree(rightInorder,rightPostorder);
45
46 return root;
47 }
48 /**
49 * Given preorder and inorder traversal of a tree,
50 * construct the binary tree.
51 * @param preorder
52 * @param inorder
53 * @return
54 */
55 public TreeNode buildTree2(int[] preorder, int[] inorder){
56 if(preorder == null || inorder == null ||
57 preorder.length == 0 || inorder.length == 0)
58 return null;
59 if(preorder.length != inorder.length)
60 return null;
61 int len = 0;
62 while(inorder[len++]!=preorder[0]);
63 //determine the left subtree
64 int[] leftInorder = new int[len-1];
65 int[] leftPreorder = new int[len-1];
66
67 int j =0;
68 while(j<len-1)
69 {
70 leftInorder[j] = inorder[j];
71 leftPreorder[j] = preorder[j+1];
72 j++;
73 }
74
75 TreeNode left = buildTree( leftPreorder,leftInorder);
76 TreeNode root = new TreeNode(preorder[0]);
77 root.left =left;
78 //determine the right subtree
79 int[] rightInorder = new int[inorder.length - (len-1) -1];
80 int[] rightPreorder = new int[preorder.length - (len-1) -1];
81 j = len;
82 while(j<inorder.length){
83 rightInorder[j-len] = inorder[j];
84 rightPreorder[j-len] = preorder[j];
85 j++;
86 }
87 root.right = buildTree(rightPreorder,rightInorder);
88 return root;
89 }
LeetCode OJ - construct Binary Tree from Inorder and
Postorder/Preorder Traversal,布布扣,bubuko.com
LeetCode OJ - construct Binary Tree from Inorder and
Postorder/Preorder Traversal
时间： 2024-08-02 07:01:30