题目
分析
深搜（一）精简
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 == 0; return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right,sum-root->val); } };
深搜（二）递归函数没有返回值，结果保存在全局变量中 !!!!
1 class Solution { 2 public: 3 bool flag = false; 4 void dfs(TreeNode*root,int t){ 5 if(root ==NULL ) return; 6 if(root->left==NULL && root->right==NULL){ //搜索到叶子结点 7 if(t == 0) {flag = true; return;} 8 } 9 if(root->left) dfs(root->left,t-root->left->val); 10 if(root->right) dfs(root->right,t-root->right->val); 11 } 12 bool hasPathSum(TreeNode* root, int targetSum) { 13 if(root == NULL) return false; 14 dfs(root,targetSum - root->val); 15 return flag; 16 } 17 };
深搜（三）递归函数存在返回值
1 class Solution { 2 public: 3 bool dfs(TreeNode* root,int t){ 4 //搜索到叶子结点 5 if(!root->left && !root->right && t == 0) return true; 6 if(!root->left && !root->right && t != 0) return false; 7 //搜索分支 8 if(root->left){ 9 if(dfs(root->left,t-root->left->val)) //包含着回溯t-root->left->val 10 return true; 11 } 12 if(root->right){ 13 if(dfs(root->right,t-root->right->val)) //包含着回溯t-root->right->val 14 return true; 15 } 16 return false; 17 } 18 bool hasPathSum(TreeNode* root, int targetSum) { 19 if(root == NULL) return false; 20 return dfs(root,targetSum - root->val); 21 } 22 };
广搜,建立两个队列，一个队列保存结点，一个队列保存结点之和
class Solution { public: bool hasPathSum(TreeNode* root, int sum) { if(root == NULL) return false; queue<TreeNode*>queue1; queue<int>queue2; queue1.push(root); queue2.push(root->val); while(!queue1.empty()){ TreeNode* node = queue1.front();queue1.pop(); int value = queue2.front();queue2.pop(); if(node->left == NULL && node->right == NULL){ if(value == sum) return true; } if(node->left != NULL){queue1.push(node->left);queue2.push(node->left->val + value);} if(node->right != NULL){queue1.push(node->right);queue2.push(node->right->val + value);} } return false; } };