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法律学C++ 程序用以执行给定二叉树的非递归中序遍历
如果一个二叉树采用中序遍历，则先访问左子树，然后访问根节点，最后访问右子树。中序遍历输出各个键的升序值。这是用于执行无递归中序树遍历的 C++ 程序。
算法
Begin Declare a structure n. Declare d of the integer datatype. Declare a pointer l against structure n. Declare a pointer r against structure n. Declare a constructor of structure n. Pass an integer variable d to parameter. this->d = d l = r = NULL Declare inOrder(struct n *root) function. Declare a stack s. Declare a pointer current against structure n. Initialize n *current = root. while (current != NULL || s.empty() == false) while (current != NULL) s.push(current) current = current->l current = s.top() s.pop() print current->d. current = current->r. insert values in nodes of tree. Call inOrder(root) function to travern the tree. End.
示例
#include<bits/stdc++.h>
using namespace std;
struct n {
int d;
struct n* l;
struct n* r;
n (int d) {
this->d = d;
l = r = NULL;
}
};
void inOrder(struct n *root) {
stack<n *> s;
n *current = root;
while (current != NULL || s.empty() == false) {
while (current != NULL) {
s.push(current);
current = current->l;
}
current = s.top();
s.pop();
cout << current->d << " ";
current = current->r;
}
}
int main() {
struct n* root = new n(6);
root->l = new n(4);
root->r= new n(7);
root->l->l = new n(8);
root->l->r= new n(5);
root->r->l = new n(9);
root->r->r = new n(10);
inOrder(root);
return 0;
}输出
8 4 5 6 9 7 10
更新日期：2019 年 7 月 30 日
614 次浏览
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