Java for LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal

標籤：

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

Note:
You may assume that duplicates do not exist in the tree.

解題思路一：

preorder[0]為root，以此分別劃分出inorderLeft、preorderLeft、inorderRight、preorderRight四個數組，然後root.left=buildTree(preorderLeft,inorderLeft); root.right=buildTree(preorderRight,inorderRight)

JAVA實現如下：

public TreeNode buildTree(int[] preorder, int[] inorder) {if (preorder.length == 0 || preorder.length != inorder.length)return null;TreeNode root = new TreeNode(preorder[0]);int index = 0;for (int i = 0; i < inorder.length; i++)if (inorder[i] == root.val) {index = 0;break;}int[] inorderLeft = new int[index], preorderLeft = new int[index];int[] inorderRight = new int[inorder.length - 1 - index], preorderRight = new int[inorder.length- 1 - index];Set<Integer> inorderLeftSet=new HashSet<Integer>();for (int i = 0; i < inorderLeft.length; i++){inorderLeft[i] = inorder[i];inorderLeftSet.add(inorder[i]);}for (int i = 0; i < inorderRight.length; i++)inorderRight[i] = inorder[index + i + 1];int j = 0, k = 0;for (int i = 0; i < preorder.length; i++) {if(inorderLeftSet.contains(preorder[i]))preorderLeft[j++]=preorder[i];else if(preorder[i]!=root.val)preorderRight[k++]=preorder[i];}if(buildTree(preorderLeft,inorderLeft)!=null)    root.left=buildTree(preorderLeft,inorderLeft);if(buildTree(preorderRight,inorderRight)!=null)    root.right=buildTree(preorderRight,inorderRight);return root;}

 結果：Time Limit Exceeded

解題思路二：

 

Java for LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal