03-树3. Tree Traversals Again (25)将先序遍历和中序遍历转为后序遍历
03-树3. Tree Traversals Again (25)
题目来源:http://www.patest.cn/contests/mooc-ds/03-%E6%A0%913
An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.
Figure 1
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.
Output Specification:
For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:6 Push 1 Push 2 Push 3 Pop Pop Push 4 Pop Pop Push 5 Push 6 Pop PopSample Output:
3 4 2 6 5 1
题目实质是通过先序遍历和中序遍历建树,再后序遍历树。
解题思路
1. 通过输入建树
Push操作代表新建一个节点,将其与父节点连接并同时压栈
Pop操作,从栈顶弹出一个节点
2. 后序遍历:递归实现
代码如下:
#include <cstdio> #include <cstring> #include <cstdlib> #define STR_LEN 5 #define MAX_SIZE 30 typedef struct Node { int data; struct Node *left, *right; }* treeNode; treeNode Stack[MAX_SIZE]; int values[MAX_SIZE]; int num = 0; int top = -1; void Push(treeNode tn); treeNode Pop(); treeNode Top(); bool isEmpty(); void PostOrderTraversal(treeNode root); int main() { int n; char operation[STR_LEN]; treeNode father, root; bool findRoot = 0, Poped = 0; scanf("%d", &n); for (int i = 0; i < 2 * n; i++) { scanf("%s", operation); if (strcmp(operation, "Push") == 0) { int value; scanf("%d", &value); treeNode newNode; newNode = (treeNode)malloc(sizeof(struct Node)); newNode->data = value; newNode->left = NULL; newNode->right = NULL; if (!findRoot) { root = newNode; //根节点 Push(newNode); findRoot = 1; } else { if (!Poped) //如果前一个操作不是pop,则父节点为栈顶元素 father = Top(); if (father->left == NULL) father->left = newNode; else father->right = newNode; //printf("%d\n", newNode->data); Push(newNode); } Poped = 0; } else { father = Pop(); Poped = 1; } } PostOrderTraversal(root); for (int i = 0; i < num-1; i++) printf("%d ", values[i]); printf("%d\n", values[num-1]); return 0; } void PostOrderTraversal(treeNode root) { treeNode tn = root; if(tn) { PostOrderTraversal(tn->left); PostOrderTraversal(tn->right); values[num++] = tn->data; //将后序遍历出的节点值存入数组便于格式化打印 } } void Push(treeNode tn) { Stack[++top] = tn; } treeNode Pop() { return Stack[top--]; } bool isEmpty() { return top == -1; } treeNode Top() { return Stack[top]; }