重建二叉树

题目描述

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并输出它的后序遍历序列。(测试用例中，"树"的输出形式类似于树的层次遍历，没有节点的用#来代替)

/**
 * 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) {
    	if (pre==null || in==null || in.length<=0) {
    		return null;
    	}
    	return ConstructCore(pre, 0, in.length-1, in, 0, in.length-1);
    }
    public static TreeNode ConstructCore(int[] preOrder,
            int startPreIndex, int endPreIndex, int[] inOrder,
            int startInIndex, int endInIndex) {

        int rootValue = preOrder[startPreIndex];
        TreeNode root = new TreeNode(rootValue);

        if (startPreIndex == endPreIndex) {
            if (startInIndex == endInIndex
                    && preOrder[startPreIndex] == inOrder[startInIndex]) {
                return root;
            }
        }
        int rootInIndex = startInIndex;

        while (rootInIndex <= endInIndex && inOrder[rootInIndex] != rootValue) {
            ++rootInIndex;
        }

        int leftLength = rootInIndex - startInIndex;

        int leftPreOrderEndIndex = startPreIndex + leftLength;

        if (leftLength > 0) {
            root.left = ConstructCore(preOrder, startPreIndex + 1,
                    leftPreOrderEndIndex, inOrder, startInIndex,
                    rootInIndex - 1);
        }

        if (leftLength < endPreIndex - startPreIndex) {
            root.right = ConstructCore(preOrder, leftPreOrderEndIndex + 1,
                    endPreIndex, inOrder, rootInIndex + 1, endInIndex);
        }
        return root;
    }
}