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 }