106.Construct Binary Tree from Inorder and Postorder Traversal

题目链接

题目大意：根据中序和后序构建二叉树。

法一：与105类似，只是这里是根据后序来确定根节点。代码如下（耗时15ms）：

 1     public TreeNode buildTree(int[] inorder, int[] postorder) {
 2         if(postorder.length == 0 || inorder.length == 0) {
 3             return null;
 4         }
 5         return dfs(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);
 6     }
 7     private TreeNode dfs(int[] inorder, int[] postorder, int inL, int inR, int postL, int postR) {
 8         TreeNode root = new TreeNode(postorder[postR]);
 9         int rootIndex = inL;
10         while(inorder[rootIndex] != postorder[postR]) {
11             rootIndex++;
12         }
13         int leftLen = rootIndex - inL;
14         int rightLen = inR - rootIndex;
15         if(leftLen != 0) {
16             root.left = dfs(inorder, postorder, inL, rootIndex - 1, postL, postL + leftLen - 1);
17         }
18         else {
19             root.left = null;
20         }
21         if(rightLen != 0) {
22             root.right = dfs(inorder, postorder, rootIndex + 1, inR, postL + leftLen, postR - 1);
23         }
24         else {
25             root.right = null;
26         }
27         return root;
28     }

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