标签：
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
解题思路一：
preorder[0]为root，以此分别划分出inorderLeft、preorderLeft、inorderRight、preorderRight四个数组，然后root.left=buildTree(preorderLeft,inorderLeft); root.right=buildTree(preorderRight,inorderRight)
JAVA实现如下：
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;
}
结果：Time Limit Exceeded
解题思路二：
Java for LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal
标签：
原文地址：http://www.cnblogs.com/tonyluis/p/4523821.html
