1020. Tree Traversals (25)
Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:7 2 3 1 5 7 6 4 1 2 3 4 5 6 7Sample Output:
4 1 6 3 5 7 2
1,根据后序遍历可以确定根结点,知道根结点后,根据中序遍历可以知道左右子树元素的分割线;
2,根据第一点,用递归的方法创建一棵树,然后用二叉树层次遍历的方法进行遍历.
#include <stdio.h>
#include <stdlib.h>struct tree_node
{int data;struct tree_node *left;struct tree_node *right;
};struct tree_node *buildTree(int *af, int afl, int afr, int *in, int inl, int inr)
{struct tree_node *ptr;int pos = 0;if (afl > afr)return NULL;else{ptr = (struct tree_node *)malloc(sizeof(struct tree_node));ptr->data = af[afr];pos = inl;while (*(in+pos) != ptr->data)pos++;ptr->left = buildTree(af, afl, afr-inr+pos-1, in, inl, pos-1);ptr->right = buildTree(af,afr-inr+pos, afr-1, in, pos+1, inr);return ptr;}
}void printfLevelOrder(struct tree_node *root)
{struct tree_node qu[100];int front, rear;front = rear = -1;if (!root)return;rear++;qu[rear].data = root->data;qu[rear].left = root->left;qu[rear].right = root->right;while (rear != front){if (front != -1)printf(" ");printf("%d", qu[++front].data);if (qu[front].left){ rear++;qu[rear].data = qu[front].left->data;qu[rear].left = qu[front].left->left;qu[rear].right = qu[front].left->right;}if (qu[front].right){ rear++;qu[rear].data = qu[front].right->data;qu[rear].left = qu[front].right->left;qu[rear].right = qu[front].right->right;}}printf("\n");
}int main()
{int N;int af[128];int in[128];int i;struct tree_node *root;scanf("%d", &N);for (i = 0; i < N; i++)scanf("%d", af+i);for (i = 0; i < N; i++)scanf("%d", in+i);root = buildTree(af, 0, N-1, in, 0, N-1);printfLevelOrder(root);return 0;
}代码说明:
第26,27行代码中
ptr->left = buildTree(af, afl, afr-inr+pos-1, in, inl, pos-1);ptr->right = buildTree(af,afr-inr+pos, afr-1, in, pos+1, inr);刚开始写成如下,总是报段错误:
ptr->left = buildTree(af, afl, pos-1, in, inl, pos-1);ptr->right = buildTree(af,pos, afr-1, in, pos+1, inr);
| index | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| af | 2 | 3 | 1 | 5 | 7 | 6 | 4 |
| in | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
观察5,7,6三个元素,就会发现第二种写法是错误的,改正的办法是怎样把上下两行5,7,6对不齐的情况考虑进去,因为pos根据in这个数字的下标得到的数子,所以pos应用到af数组时,把偏移量afr-inr=-1加上.
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