原文:
给定一棵二叉树，任务是从左到右打印给定二叉树的所有叶节点。也就是说，节点应该按照它们在给定树中从左到右的顺序打印。 例:
input :
1
/ \
2 3
/ \ / \
4 5 6 7
output : 4 5 6 7
input :
4
/ \
5 9
/ \ / \
8 3 7 2
/ / \
12 6 1
output : 12 3 7 6 1
我们已经讨论了使用两个堆栈的方法。 方法:想法是使用一个堆栈执行迭代
c
// c program to print leaf nodes from
// left to right using one stack
#include
using namespace std;
// structure of binary tree
struct node {
node* left;
node* right;
int data;
};
// function to create a new node
node* newnode(int key)
{
node* node = new node();
node->left = node->right = null;
node->data = key;
return node;
}
// function to print all the leaf nodes
// of binary tree using one stack
void printleaflefttoright(node* p)
{
// stack to store the nodes
stack s;
while (1) {
// if p is not null then push
// it on the stack
if (p) {
s.push(p);
p = p->left;
}
else {
// if stack is empty then come out
// of the loop
if (s.empty())
break;
else {
// if the node on top of the stack has its
// right subtree as null then pop that node and
// print the node only if its left
// subtree is also null
if (s.top()->right == null) {
p = s.top();
s.pop();
// print the leaf node
if (p->left == null)
printf("%d ", p->data);
}
while (p == s.top()->right) {
p = s.top();
s.pop();
if (s.empty())
break;
}
// if stack is not empty then assign p as
// the stack's top node's right child
if (!s.empty())
p = s.top()->right;
else
p = null;
}
}
}
}
// driver code
int main()
{
node* root = newnode(1);
root->left = newnode(2);
root->right = newnode(3);
root->left->left = newnode(4);
root->left->right = newnode(5);
root->right->left = newnode(6);
root->right->right = newnode(7);
printleaflefttoright(root);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to print leaf nodes from
// left to{ right using one stack
import java.util.*;
class gfg
{
// structure of binary tree
static class node
{
node left;
node right;
int data;
}
// function to create a new node
static node newnode(int key)
{
node node = new node();
node.left = null;
node.right = null;
node.data = key;
return node;
}
// function to print all the leaf nodes
// of binary tree using one stack
static void printleaflefttoright(node p)
{
// stack to store the nodes
stack s = new stack ();
while (true)
{
// if p is not null then push
// it on the stack
if (p != null)
{
s.push(p);
p = p.left;
}
else
{
// if stack is empty then come out
// of the loop
if (s.isempty())
break;
else
{
// if the node on top of the stack has its
// right subtree as null then pop that node and
// print the node only if its left
// subtree is also null
if (s.peek().right == null)
{
p = s.peek();
s.pop();
// print the leaf node
if (p.left == null)
system.out.print(p.data " ");
}
while (p == s.peek().right)
{
p = s.peek();
s.pop();
if (s.isempty())
break;
}
// if stack is not empty then assign p as
// the stack's top node's right child
if (!s.isempty())
p = s.peek().right;
else
p = null;
}
}
}
}
// driver code
public static void main(string[] args)
{
node root = newnode(1);
root.left = newnode(2);
root.right = newnode(3);
root.left.left = newnode(4);
root.left.right = newnode(5);
root.right.left = newnode(6);
root.right.right = newnode(7);
printleaflefttoright(root);
}
}
// this code is contributed by prerna saini
python 3
# python3 program to print leaf nodes from
# left to right using one stack
# binary tree node
class newnode:
def __init__(self, data):
self.data = data
self.left = none
self.right = none
# function to print all the leaf nodes
# of binary tree using one stack
def printleaflefttoright(p):
# stack to store the nodes
s = []
while (1):
# if p is not none then push
# it on the stack
if (p):
s.insert(0, p)
p = p.left
else:
# if stack is empty then come out
# of the loop
if len(s) == 0:
break
else:
# if the node on top of the stack has its
# right subtree as none then pop that node
# and print the node only if its left
# subtree is also none
if (s[0].right == none):
p = s[0]
s.pop(0)
# print the leaf node
if (p.left == none):
print(p.data, end = " ")
while (p == s[0].right):
p = s[0]
s.pop(0)
if len(s) == 0:
break
# if stack is not empty then assign p as
# the stack's top node's right child
if len(s):
p = s[0].right
else:
p = none
# driver code
root = newnode(1)
root.left = newnode(2)
root.right = newnode(3)
root.left.left = newnode(4)
root.left.right = newnode(5)
root.right.left = newnode(6)
root.right.right = newnode(7)
printleaflefttoright(root)
# this code is contributed by shubhamsingh10
c
// c# program to print leaf nodes from
// left to{ right using one stack
using system;
using system.collections.generic;
class gfg
{
// structure of binary tree
public class node
{
public node left;
public node right;
public int data;
}
// function to create a new node
static node newnode(int key)
{
node node = new node();
node.left = null;
node.right = null;
node.data = key;
return node;
}
// function to print all the leaf nodes
// of binary tree using one stack
static void printleaflefttoright(node p)
{
// stack to store the nodes
stack s = new stack ();
while (true)
{
// if p is not null then push
// it on the stack
if (p != null)
{
s.push(p);
p = p.left;
}
else
{
// if stack is empty then come out
// of the loop
if (s.count == 0)
break;
else
{
// if the node on top of the stack has its
// right subtree as null then pop that node and
// print the node only if its left
// subtree is also null
if (s.peek().right == null)
{
p = s.peek();
s.pop();
// print the leaf node
if (p.left == null)
console.write(p.data " ");
}
while (p == s.peek().right)
{
p = s.peek();
s.pop();
if (s.count == 0)
break;
}
// if stack is not empty then assign p as
// the stack's top node's right child
if (s.count != 0)
p = s.peek().right;
else
p = null;
}
}
}
}
// driver code
public static void main(string[] args)
{
node root = newnode(1);
root.left = newnode(2);
root.right = newnode(3);
root.left.left = newnode(4);
root.left.right = newnode(5);
root.right.left = newnode(6);
root.right.right = newnode(7);
printleaflefttoright(root);
}
}
// this code has been contributed by 29ajaykumar
java 描述语言
output:
4 5 6 7
时间复杂度:o(n) t3】辅助空间: o(n)
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