重建二叉树

 

/**
 * 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 t = refuction(pre,0,pre.length-1,in,0,in.length-1);
        return t;
        }
    private TreeNode refuction(int [] pre, int preStart,int preEnd,int [] in,int inStart,int inEnd){
        if (preStart > preEnd || inStart > inEnd) return null;
        TreeNode root = new TreeNode(pre[preStart]);
        for(int i = inStart;i <= inEnd;i++){
            if(in[i] == pre[preStart]){
//前序遍历首个节点为二叉树的根节点，找到在中序遍历根节点的位置，则左边为左子树，右边为右子树。
                root.left = refuction(pre,preStart+1,preStart+(i-inStart),in,inStart,i-1 );
                root.right = refuction(pre,preStart+(i-inStart)+1,preEnd,in,i+1,inEnd);
                break;
            }
        }
        return root;
    }
}