原文:
给定一棵二叉树和一个和 s ，打印所有路径，从根开始，求和到给定的和。 注意，这个问题与不同。这里的路径不需要在叶节点上结束。
示例:
input :
input : sum = 8,
root of tree
1
/ \
20 3
/ \
4 15
/ \ / \
6 7 8 9
output :
path: 1 3 4
input : sum = 38,
root of tree
10
/ \
28 13
/ \
14 15
/ \ / \
21 22 23 24
output : path found: 10 28
path found: 10 13 15
对于这个问题，前序遍历是最适合的，因为我们必须在到达那个节点时添加一个键值。 我们从根开始，通过前序遍历开始遍历，给 sum_so_far 加上键值，检查是否等于需要的和。 同样，由于树节点没有指向其父节点的指针，我们必须从我们移动的地方显式保存。我们使用向量路径来存储该路径。 该路径中的每个节点都有助于 sum_so_far。
c
// c program to print all paths begiinning with
// root and sum as given sum
#include
using namespace std;
// a tree node
struct node
{
int key;
struct node *left, *right;
};
// utility function to create a new node
node* newnode(int key)
{
node* temp = new node;
temp->key = key;
temp->left = temp->right = null;
return (temp);
}
void printpathsutil(node* curr_node, int sum,
int sum_so_far, vector &path)
{
if (curr_node == null)
return;
// add the current node's value
sum_so_far = curr_node->key;
// add the value to path
path.push_back(curr_node->key);
// print the required path
if (sum_so_far == sum )
{
cout << "path found: ";
for (int i=0; ileft != null)
printpathsutil(curr_node->left, sum, sum_so_far, path);
// if right child exists
if (curr_node->right != null)
printpathsutil(curr_node->right, sum, sum_so_far, path);
// remove last element from path
// and move back to parent
path.pop_back();
}
// wrapper over printpathsutil
void printpaths(node *root, int sum)
{
vector path;
printpathsutil(root, sum, 0, path);
}
// driver program
int main ()
{
/* 10
/ \
28 13
/ \
14 15
/ \ / \
21 22 23 24*/
node *root = newnode(10);
root->left = newnode(28);
root->right = newnode(13);
root->right->left = newnode(14);
root->right->right = newnode(15);
root->right->left->left = newnode(21);
root->right->left->right = newnode(22);
root->right->right->left = newnode(23);
root->right->right->right = newnode(24);
int sum = 38;
printpaths(root, sum);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to print all paths beginning
// with root and sum as given sum
import java.util.arraylist;
class graph{
// a tree node
static class node
{
int key;
node left, right;
};
// utility function to create a new node
static node newnode(int key)
{
node temp = new node();
temp.key = key;
temp.left = temp.right = null;
return (temp);
}
static void printpathsutil(node curr_node, int sum,
int sum_so_far,
arraylist path)
{
if (curr_node == null)
return;
// add the current node's value
sum_so_far = curr_node.key;
// add the value to path
path.add(curr_node.key);
// print the required path
if (sum_so_far == sum)
{
system.out.print("path found: ");
for(int i = 0; i < path.size(); i )
system.out.print(path.get(i) " ");
system.out.println();
}
// if left child exists
if (curr_node.left != null)
printpathsutil(curr_node.left, sum,
sum_so_far, path);
// if right child exists
if (curr_node.right != null)
printpathsutil(curr_node.right, sum,
sum_so_far, path);
// remove last element from path
// and move back to parent
path.remove(path.size() - 1);
}
// wrapper over printpathsutil
static void printpaths(node root, int sum)
{
arraylist path = new arraylist<>();
printpathsutil(root, sum, 0, path);
}
// driver code
public static void main(string[] args)
{
/* 10
/ \
28 13
/ \
14 15
/ \ / \
21 22 23 24*/
node root = newnode(10);
root.left = newnode(28);
root.right = newnode(13);
root.right.left = newnode(14);
root.right.right = newnode(15);
root.right.left.left = newnode(21);
root.right.left.right = newnode(22);
root.right.right.left = newnode(23);
root.right.right.right = newnode(24);
int sum = 38;
printpaths(root, sum);
}
}
// this code is contributed by sanjeev2552
python 3
# python3 program to print all the
# paths from root, with a specified
# sum in binary tree
# binary tree node
""" utility that allocates a newnode
with the given key """
class newnode:
# construct to create a newnode
def __init__(self, key):
self.key = key
self.left = none
self.right = none
# this function prints all paths
# that have sum k
def printpathsutil(curr_node, sum,
sum_so_far, path):
# empty node
if (not curr_node) :
return
sum_so_far = curr_node.key
# add current node to the path
path.append(curr_node.key)
# print the required path
if (sum_so_far == sum ) :
print("path found:", end = " ")
for i in range(len(path)):
print(path[i], end = " ")
print()
# if left child exists
if (curr_node.left != none) :
printpathsutil(curr_node.left, sum,
sum_so_far, path)
# if right child exists
if (curr_node.right != none) :
printpathsutil(curr_node.right, sum,
sum_so_far, path)
# remove the current element
# from the path
path.pop(-1)
# a wrapper over printkpathutil()
def printpaths(root, sum):
path = []
printpathsutil(root, sum, 0, path)
# driver code
if __name__ == '__main__':
""" 10
/ \
28 13
/ \
14 15
/ \ / \
21 22 23 24"""
root = newnode(10)
root.left = newnode(28)
root.right = newnode(13)
root.right.left = newnode(14)
root.right.right = newnode(15)
root.right.left.left = newnode(21)
root.right.left.right = newnode(22)
root.right.right.left = newnode(23)
root.right.right.right = newnode(24)
sum = 38
printpaths(root, sum)
# this code is contributed by
# shubham singh(shubhamsingh10)
java 描述语言
输出:
path found: 10 28
path found: 10 13 15
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