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

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：出现上次解法的原因是因为本人把前序遍历理解成了层次遍历，其实本题是《编程之美》3.9节 重建二叉树的原题，书中已经给出的答案，JAVA实现如下：

	static public TreeNode buildTree(int[] preorder, int[] inorder) {
		 return buildTree(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);  
    }  
    static public TreeNode buildTree(int[] preorder, int[] inorder, int pBegin, int pEnd, int iBegin, int iEnd){  
        if(pBegin>pEnd)  
            return null;  
        TreeNode root = new TreeNode(preorder[pBegin]);  
        int i = iBegin;  
        for(;i<iEnd;i++)
            if(inorder[i]==root.val)  
                break;  
        int lenLeft = i-iBegin;  
        root.left = buildTree(preorder, inorder, pBegin+1, pBegin+lenLeft, iBegin, i-1);  
        root.right = buildTree(preorder, inorder, pBegin+lenLeft+1, pEnd, i+1, iEnd);  
        return root;
    }