Traversal de límite del árbol binario
#include <bits/stdc++.h>
using namespace std;
struct node {
int data;
struct node * left, * right;
};
bool isLeaf(node * root) {
return !root -> left && !root -> right;
}
void addLeftBoundary(node * root, vector < int > & res) {
node * cur = root -> left;
while (cur) {
if (!isLeaf(cur)) res.push_back(cur -> data);
if (cur -> left) cur = cur -> left;
else cur = cur -> right;
}
}
void addRightBoundary(node * root, vector < int > & res) {
node * cur = root -> right;
vector < int > tmp;
while (cur) {
if (!isLeaf(cur)) tmp.push_back(cur -> data);
if (cur -> right) cur = cur -> right;
else cur = cur -> left;
}
for (int i = tmp.size() - 1; i >= 0; --i) {
res.push_back(tmp[i]);
}
}
void addLeaves(node * root, vector < int > & res) {
if (isLeaf(root)) {
res.push_back(root -> data);
return;
}
if (root -> left) addLeaves(root -> left, res);
if (root -> right) addLeaves(root -> right, res);
}
vector < int > printBoundary(node * root) {
vector < int > res;
if (!root) return res;
if (!isLeaf(root)) res.push_back(root -> data);
addLeftBoundary(root, res);
// add leaf nodes
addLeaves(root, res);
addRightBoundary(root, res);
return res;
}
struct node * newNode(int data) {
struct node * node = (struct node * ) malloc(sizeof(struct node));
node -> data = data;
node -> left = NULL;
node -> right = NULL;
return (node);
}
int main() {
struct node * root = newNode(1);
root -> left = newNode(2);
root -> left -> left = newNode(3);
root -> left -> left -> right = newNode(4);
root -> left -> left -> right -> left = newNode(5);
root -> left -> left -> right -> right = newNode(6);
root -> right = newNode(7);
root -> right -> right = newNode(8);
root -> right -> right -> left = newNode(9);
root -> right -> right -> left -> left = newNode(10);
root -> right -> right -> left -> right = newNode(11);
vector < int > boundaryTraversal;
boundaryTraversal = printBoundary(root);
cout << "The Boundary Traversal is : ";
for (int i = 0; i < boundaryTraversal.size(); i++) {
cout << boundaryTraversal[i] << " ";
}
return 0;
}
Different Dotterel