Reconstruction of a binary tree to prove safety Offer-

Subject 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.

 

 

public class Solution {
    class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }
    }


    public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
        if (pre != null && in != null)
            return solution(pre, in, 0, pre.length - 1, 0, in.length - 1);
        else
            return null;
    }

    private TreeNode solution(int[] pre, int[] in, int pStart, int pEnd, int iStart, int iEnd) {

        if (pStart > pEnd || iStart > iEnd)
            return null;

        TreeNode tree = new TreeNode(pre[pStart]);

        for (int root = iStart; root <= iEnd; root++)
            if (in[root] == pre[pStart]) {
                int leftLen = root - iStart;
                tree.left = solution(pre, in, pStart + 1, pStart + leftLen, iStart, root - 1);
                tree.right = solution(pre, in, pStart + leftLen + 1, pEnd, root + 1, iEnd);
                break;
            }
        return tree;
    }
}