根据遍历序列重建二叉树

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}，则重建二叉树并返回。

1）根据前序、中序遍历重建二叉树

/**
 * 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 = createTree(pre, in, 0, 0, in.length);
        
        return t;
    }
	
	public static TreeNode createTree ( int [] pre, int [] in, int i, int low, int high) {
		// i 为当前的树根节点在前序遍历数组的位置
		// [low, high) 表示当前处理的树所有节点在中序数组中的分布范围
		TreeNode t = new TreeNode(pre[i]);
		// index 是当前树的根节点在中序遍历数组中的位置
		
		int index = low;
		for ( ; index < high; index++ ) {
			if ( in[index] == pre[i] ) 
				break;
		}
		
		if( index - 1 >= low ) {
			t.left = createTree(pre, in, i+1, low, index);
		}
		else {
			t.left = null;
		}
		
		if ( index + 1 < high) {
			t.right = createTree(pre, in, i + index - low + 1, index + 1, high);
		}
		else {
			t.right = null;
		}
		
		return t;
	}
}

2） 根据后序遍历和中序遍历重建二叉树，原理差不多

BTree createTree2 ( int *post, int *in, int i, int low, int high ) {
    int index;
    BTree t = (BTree)malloc(sizeof(BinaryTree));
    t->value = post[i];

    index = low;
    for ( ; index < high; index++ ) {
        if ( in[index] == post[i] ) {
            break;
        }
    }

    if ( index - 1 >= low ) {
        t->leftChild = createTree2(post, in, i - high + index, low, index);
    }
    else {
        t->leftChild = NULL;
    }

    if ( index + 1 < high ) {
        t->rightChild = createTree2(post, in, i - 1, index + 1, high);
    }
    else {
        t->rightChild = NULL;
    }

    return t;
}

BTree reConstructBTree2 ( int *post, int *in, int n ) {
    return createTree2(post, in, n - 1, 0, n);
}