PAT 1151 LCA in a Binary Tree[難][二叉樹]
1151 LCA in a Binary Tree (30 分)
The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.
Given any two nodes in a binary tree, you are supposed to find their LCA.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), the number of keys in the binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively. It is guaranteed that the binary tree can be uniquely determined by the input sequences. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.
Output Specification:
For each given pair of U and V, print in a line LCA of U and V is A.
if the LCA is found and A
is the key. But if A
is one of U and V, print X is an ancestor of Y.
where X
is A
and Y
is the other node. If U or V is not found in the binary tree, print in a line ERROR: U is not found.
ERROR: V is not found.
or ERROR: U and V are not found.
.
Sample Input:
6 8
7 2 3 4 6 5 1 8
5 3 7 2 6 4 8 1
2 6
8 1
7 9
12 -3
0 8
99 99
Sample Output:
LCA of 2 and 6 is 3. 8 is an ancestor of 1. ERROR: 9 is not found. ERROR: 12 and -3 are not found. ERROR: 0 is not found. ERROR: 99 and 99 are not found.
題目大意:給出一個二叉樹的中根遍歷和前根遍歷,然後給出查詢,查詢這兩個節點的最低公共子節點。
//當時看到題目之後,就想到了之前有一道類似的,不過那個二叉樹是二叉搜索樹。
//使用前根和中根可以確定後根遍歷結果,不用重建二叉樹,然後根據前根和後根,假設查詢節點為u和v
//比如前根和後跟遍歷中,既在u/v節點的前面又在其後面,那麽肯定就是u/v的祖父節點,然後比較二者的祖父節點,查看最低相同的。
//當時的思路是這樣,不過只得了22分,有一個測試點是超時的,不知道怎麽改。
代碼來自:https://www.liuchuo.net/archives/6496
PAT 1151 LCA in a Binary Tree[難][二叉樹]