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Leetcode Path Sum

2024-07-10 11:04:46   229人阅读

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,
              5             /             4   8           /   /           11  13  4         /  \              7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

递归实现

从树根开始判断

  • 如果当前是叶子节点等于剩余和，那么程序返回真
  • 如果当前节点为空，则放回false
  • 除了以上情况，返回递归调用左右子节点的结果或值，并且传入剩余和扣除当前节点值后的值
class Solution {public:    bool hasPathSum(TreeNode *root, int sum) {        if(root == NULL) return false;        if(root->left == NULL && root->right == NULL && sum == root->val) return true;        else return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right, sum-root->val);    }};
class Solution {public:    bool hasPathSum(TreeNode *root, int sum) {        if(root == NULL) return false;        if(root->left == NULL && root->right == NULL) return sum == root->val;        sum-=root->val;        return hasPathSum(root->left,sum) || hasPathSum(root->right, sum);    }};
递归实现

 非递归遍历（根据层次遍历实现即可）

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */#include <iostream>#include <queue>using namespace std;struct TreeNode{    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x): val(x), left(NULL),right(NULL){}};struct TreeNodeSum{    TreeNode *node;    int sum;    TreeNodeSum(TreeNode* node_, int sum_ ):node(node_),sum(sum_){}};bool hasPathSum(TreeNode *root, int sum) {    if(root == NULL) return false;    queue<TreeNodeSum *> que;    que.push(new TreeNodeSum(root,root->val));    int res = 0;    while(!que.empty()){        TreeNodeSum *tmp = que.front();que.pop();        TreeNode *node = tmp->node;        cout<<node->val<<endl;        if(!node->left && !node->right && tmp->sum == sum) return true;         if(node->left){            TreeNodeSum *nodeSum = new TreeNodeSum(node->left,node->left->val+tmp->sum);            que.push(nodeSum);        }        if(node->right ){            TreeNodeSum *nodeSum = new TreeNodeSum(node->right,node->right->val+tmp->sum);            que.push(nodeSum);        }    }    return false;}int main(){    TreeNode *root = new TreeNode(1);    TreeNode *root1 = new TreeNode(-2);    TreeNode *root2 = new TreeNode(-3);    TreeNode *root3 = new TreeNode(1);    TreeNode *root4 = new TreeNode(3);    TreeNode *root5 = new TreeNode(-2);    TreeNode *root6 = new TreeNode(-1);    root->left = root1;    root->right = root2;    root1->left = root3;    root1->right = root4;    root2->left = root5;    root3->left = root6;    cout<<hasPathSum(root,-1)<<endl;}
非递归实现


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