Construct Binary Tree from Preorder and Inorder Traversal

標籤：

Given preorder and inorder traversal of a tree, construct the binary tree.

這道題是要根據先序遍曆和中序遍曆構造出一棵二叉樹，這道題和上一題是一樣的。上一題不過是通過後序遍曆和中序遍曆構造一棵二叉樹。

只要將代碼稍稍修改即可

 1 public class Solution { 2     public TreeNode buildTree(int[] preorder, int[] inorder) { 3         TreeNode root; 4         if(preorder.length == 0) 5             return null; 6         return createTree(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1); 7  8     } 9     private TreeNode createTree(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart ,int inEnd){10         if(inStart > inEnd)11             return null;12         int value_root = preorder[preStart];                    //根據先序遍曆第一個元素為根節點,找出根節點的值13         int index = 0;                                            //根節點在中序遍曆中的索引值14         for(int i = inStart; i <= inEnd; i++){                    //尋找根節點在中序遍曆中的索引值15             if(value_root == inorder[i]){16                 index = i;17                 break;18             }19         }//for20         int length_leftTree = index - inStart;                    //左子樹的長度21         TreeNode root_left = createTree(preorder, inorder, preStart + 1, preStart + length_leftTree, inStart, index - 1);22         TreeNode root_right = createTree(preorder, inorder, preStart + length_leftTree + 1, preEnd, index + 1, inEnd);23         24         TreeNode root = new TreeNode(value_root);25         root.left = root_left;26         root.right = root_right;27         28         return root;29     }30 }

 

Construct Binary Tree from Preorder and Inorder Traversal