一颗二叉树前序遍历和中序遍历分别是ABDEGCFH、DBGEACHF,则此后序遍历是？请高手解释怎么得的，说明原理！
推荐回答（3个）
前序可知A是根结点，由A在中序中的位置可以看出A的左子树上包括DBGE四个结点，右子树上包括CHF三个结点。由前序列可以看出B结点是左子树的根结点，再结合B在上述四个结点中的位置可以得出B的左子树为D右子树包含GE两个结点。由前序列E在G前面，说明E是B的右结点，中序排列中G在E前面说明G是E的左结点，A的右子树也可以这样推出来。
画图最简单了，由第一序列先画上根结点A，第一数列中第二位是B，在第二个数列中B在已确定的A的左侧，那么B就是A的左结点，B也确定了。第一数列中第三位是D，D在第二数列中位于已确定的B的左侧，那D就是B的左结点；第一数列中第四个是E，E在第二个数列中已确定的B的右侧，那么E就是B的右结点；第五个是G，G在第二数列中位于已确定的E的左侧，那么G就是E的左结点；第六个是C，C在第二个数列中位于已确定点A的右侧，C是A的右结点；下一个是F，F在已确定结点C的右侧，F是C的右结点；最后一个H，H在C的右侧F的左侧，则F是C的左结点。好了整个二叉树出来了，后序遍历自己看就行了。
由前序遍历确定根的位置，中序遍历确定左右子树包括的结点，然后分成两棵子树的相同子问题来递归求解，如此即可确定树的结构，后序遍历就没问题了。
(DBGE)A(CHF)→((D)B(GE))A(C(HF)→((D)B((G)E))A(C((H)F)，
所以后序遍历为DGEBHFCA
前序就是先根，则A必定是根！！！
从中序就能从A一分为二，分别为左与右的子树；
中序： 左子中序 DBGE 右子中树 CHF
再从前序中分出左、右子树：
前序 左子前序： BDEG 右子前序 CFH
这个问题递归成两个小问题。同样的方法再分析出两个子树
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