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.
Tags:Tree Depth-first Search
Solution 1: recursion
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool hasPathSum(TreeNode* root, int sum) { if(!root)return false; if(!root->left && !root->right ){ if(root->val==sum) return true; else return false; } return hasPathSum(root->left, sum-(root->val)) || hasPathSum(root->right, sum-(root->val)); } };
Solution 2: stack
bool hasPathSum(TreeNode* root, int sum) { if(!root)return false; int cnt=root->val; stack<TreeNode*> stk; unordered_map<TreeNode*,bool> visited; stk.push(root); visited[root]=true; while(!stk.empty()){ TreeNode* top=stk.top(); if(!top->left&&!top->right){ if(cnt==sum)return true; } if(top->left&&visited[top->left]==false){ stk.push(top->left); visited[top->left]=true; cur+=top->left->val; continue; } if(top->right&&visited[top->right]==false){ stk.push(top->right); visited[top->right]=true; cur+=top->right->val; continue; } stk.pop(); cnt-=top->val; } return false; }
时间: 2024-10-06 09:30:47