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Find the sum of all left leaves in a given binary tree.
Example:
3
/ 9 20
/ 15 7
There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
Analyse: Check if the current node has left child and if the left child is a leaf node.
Runtime: 3ms
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int sumOfLeftLeaves(TreeNode* root) { 13 if (!root || (!root->left && !root->right)) return 0; 14 15 int result = 0; 16 sumLeftLeaves(root, result); 17 return result; 18 } 19 20 void sumLeftLeaves(TreeNode* root, int &result) { 21 if (root->left && !root->left->left && !root->left->right) 22 result += root->left->val; 23 24 if (root->left) sumLeftLeaves(root->left, result); 25 if (root->right) sumLeftLeaves(root->right, result); 26 } 27 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/5922453.html
