【二叉树的递归】04找出二叉树中路径和等于给定值的所有路径【Path Sum II】

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给定一个二叉树和一个和，判断这个树中是否有一个从根到叶子的路径，使其这个路径上面的所有节点值的和为这个给定的值。

并且返回所有等于给定值的路径。

例如：

给定下面的二叉树，并且和为22。

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

返回true，因为这里面存在一个根到叶子的路径 5->4->11->2，使其他们的和为22。

返回：

[
   [5,4,11,2],
   [5,8,4,5]
]

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Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

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test.cpp:

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#include <iostream>
#include <cstdio>
#include <stack>
#include <vector>
#include "BinaryTree.h"

using namespace std;

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
void findPathSum(TreeNode *root, vector< int> &path, vector<vector< int> > &allpath, int sum)
{
/*没访问一个节点就放进去*/
path.push_back(root->val);
if(root->left == NULL && root->right == NULL)
{
vector< int>::iterator it = path.begin();
int tmpsum = 0;
for(; it != path.end(); ++it)
{
tmpsum += *it;
}
if(tmpsum == sum)
{
allpath.push_back(path);
}
/*到叶子节点了，而且已经计算过了，就把这个节点退出，并且递归结束，这种节点是叶子节点的情况*/
path.pop_back();
return ;
}

if(root->left != NULL)
{
findPathSum(root->left, path, allpath, sum);
}
if(root->right != NULL)
{
findPathSum(root->right, path, allpath, sum);
}
/*如果这个节点的左右节点都访问过了，把这个节点退出，并且返回上层递归，这种节点是左右不同时空的情况的节点*/
path.pop_back();
}

vector<vector< int> > pathSum(TreeNode *root, int sum)
{

vector<vector< int> > allpath;
if(root == NULL)
{
return allpath;
}
vector< int> path;
findPathSum(root, path, allpath, sum);
return allpath;
}

// 树中结点含有分叉，
// 8
// / \
// 6 12
// / \
// 9 2
// / \
// 4 7
int main()
{
TreeNode *pNodeA1 = CreateBinaryTreeNode( 8);
TreeNode *pNodeA2 = CreateBinaryTreeNode( 6);
TreeNode *pNodeA3 = CreateBinaryTreeNode( 12);
TreeNode *pNodeA4 = CreateBinaryTreeNode( 9);
TreeNode *pNodeA5 = CreateBinaryTreeNode( 2);
TreeNode *pNodeA6 = CreateBinaryTreeNode( 4);
TreeNode *pNodeA7 = CreateBinaryTreeNode( 7);

ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);

PrintTree(pNodeA1);

vector<vector< int> > ans = pathSum(pNodeA1, 20);

for ( int i = 0; i < ans.size(); ++i)
{
for ( int j = 0; j < ans[i].size(); ++j)
{
cout << ans[i][j] << " ";
}
cout << endl;
}
cout << endl;

DestroyTree(pNodeA1);
return 0;
}

结果输出：

8 6 2 4

8 12

BinaryTree.h:

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#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_

struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode( int x) : val(x), left( NULL), right( NULL) {}
};

TreeNode *CreateBinaryTreeNode( int value);
void ConnectTreeNodes(TreeNode *pParent,
TreeNode *pLeft, TreeNode *pRight);
void PrintTreeNode(TreeNode *pNode);
void PrintTree(TreeNode *pRoot);
void DestroyTree(TreeNode *pRoot);

#endif /*_BINARY_TREE_H_*/

BinaryTree.cpp:

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#include <iostream>
#include <cstdio>
#include "BinaryTree.h"

using namespace std;

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

//创建结点
TreeNode *CreateBinaryTreeNode( int value)
{
TreeNode *pNode = new TreeNode(value);

return pNode;
}

//连接结点
void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight)
{
if(pParent != NULL)
{
pParent->left = pLeft;
pParent->right = pRight;
}
}

//打印节点内容以及左右子结点内容
void PrintTreeNode(TreeNode *pNode)
{
if(pNode != NULL)
{
printf( "value of this node is: %d\n", pNode->val);

if(pNode->left != NULL)
printf( "value of its left child is: %d.\n", pNode->left->val);
else
printf( "left child is null.\n");

if(pNode->right != NULL)
printf( "value of its right child is: %d.\n", pNode->right->val);
else
printf( "right child is null.\n");
}
else
{
printf( "this node is null.\n");
}

printf( "\n");
}

//前序遍历递归方法打印结点内容
void PrintTree(TreeNode *pRoot)
{
PrintTreeNode(pRoot);

if(pRoot != NULL)
{
if(pRoot->left != NULL)
PrintTree(pRoot->left);

if(pRoot->right != NULL)
PrintTree(pRoot->right);
}
}

void DestroyTree(TreeNode *pRoot)
{
if(pRoot != NULL)
{
TreeNode *pLeft = pRoot->left;
TreeNode *pRight = pRoot->right;

delete pRoot;
pRoot = NULL;

DestroyTree(pLeft);
DestroyTree(pRight);
}
}

转载于:https://www.cnblogs.com/codemylife/p/3652387.html

【二叉树的递归】04找出二叉树中路径和等于给定值的所有路径【Path Sum II】

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