LeetCode404.左叶子之和
题目
法一、广度优先搜索
1 class Solution { 2 public: 3 int sumOfLeftLeaves(TreeNode* root) { 4 if(root == NULL) return 0; 5 if(root->left == NULL && root->right == NULL) return 0; 6 int sum = 0; 7 queue<TreeNode*>que; 8 que.push(root); 9 while(!que.empty()){ 10 TreeNode* node = que.front();que.pop(); 11 if(node->left != NULL) { 12 que.push(node->left); 13 if(node->left->left == NULL && node->left->right == NULL) 14 sum += node->left->val; 15 } 16 if(node->right != NULL) { 17 que.push(node->right); 18 } 19 } 20 return sum; 21 } 22 };
法二、深搜
左叶子之和 = 左子树的左叶子 + 右子树的左叶子
1 class Solution { 2 public: 3 bool isLeafNode(TreeNode *root){ 4 return (!root->left && !root->right); 5 } 6 7 int dfs(TreeNode* root){ 8 int sum = 0; 9 if(root->left != NULL){ 10 if(root->left != NULL && isLeafNode(root->left)) 11 sum += root->left->val; 12 if(!isLeafNode(root->left)) 13 sum += dfs(root->left); 14 } 15 if(root->right != NULL && !isLeafNode(root->right)) 16 sum += dfs(root->right); 17 return sum; 18 } 19 20 int sumOfLeftLeaves(TreeNode* root) { 21 if(root == NULL) return 0; 22 return dfs(root); 23 } 24 25 26 };
注意 sum = 0 一定要放在 dfs内部不能作为全局变量,否则将重复加值
另外一种写法
1 class Solution { 2 public: 3 bool isLeaf(TreeNode* root){ 4 if(root!=NULL && !root->left && !root->right) return true; 5 else return false; 6 } 7 int sumOfLeftLeaves(TreeNode* root) { 8 if(root == NULL) return 0; 9 if(isLeaf(root->left)) return root->left->val + sumOfLeftLeaves(root->right); 10 else return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right); 11 } 12 };