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.
思路：
递归遍历，判断是左叶子加入sum.
void tr(TreeNode* root, int& sum) { if (root == NULL)return; if (root->left != NULL && root->left->left == NULL && root->left->right == NULL)sum += root->left->val; tr(root->left, sum); tr(root->right, sum); } int sumOfLeftLeaves(TreeNode* root) { if (root == NULL)return 0; int sum = 0; tr(root, sum); return sum; }
[leetcode-404-Sum of Left Leaves]
原文：http://www.cnblogs.com/hellowooorld/p/7117390.html