Wins the offer before the fourth preface and in order to rebuild binary tree

Title Description

And enter the result in a preorder traversal of a binary tree in preorder traversal of the binary tree a rebuild. Suppose Results preorder traversal order and input of duplicate numbers are free. Before entering e.g. preorder traversal sequence {1,2,4,7,3,5,6,8} and {4,7,2,1,5,3,8,6} order traversal sequence, and the reconstructed binary tree return.

This problem is very ashamed, copy someone else's, the big brother Code stick out, could not find a big brother blog (so do not stick indexes, seeking forgive me, big brother can contact me at

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
    public class Solution {
	public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
		TreeNode root = reConstructBinaryTree(pre, 0, pre.length - 1, in, 0, in.length - 1);
		return root;
	}

	// 前序遍历{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}
	private TreeNode reConstructBinaryTree(int[] pre, int startPre, int endPre, int[] in, int startIn, int endIn) {

		if (startPre > endPre || startIn > endIn) {
			return null;
		}
		TreeNode root = new TreeNode(pre[startPre]);
		for (int i = startIn; i <= endIn; i++)
			if (in[i] == pre[startPre]) {
				root.left = reConstructBinaryTree(pre, startPre + 1, startPre + i - startIn, in, startIn, i - 1);
				root.right = reConstructBinaryTree(pre, i - startIn + startPre + 1, endPre, in, i + 1, endIn);
				break;
			}
		return root;
	}
}

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