Find sum of all left leaves in a given Binary Tree - GeeksforGeeks
Given a Binary Tree, find sum of all left leaves in it. For example, sum of all left leaves in below Binary Tree is 5+23+50 = 78.
O(N)
This solution passes in a sum variable as an accumulator. When a left leaf is encountered, the leaf’s data is added to sum.
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Given a Binary Tree, find sum of all left leaves in it. For example, sum of all left leaves in below Binary Tree is 5+23+50 = 78.
O(N)
bool isLeaf(Node *node){ if (node == NULL) return false; if (node->left == NULL && node->right == NULL) return true; return false;}// This function returns sum of all left leaves in a given// binary treeint leftLeavesSum(Node *root){ int res = 0; // Update result if root is not NULL if (root != NULL) { // If left of root is NULL, then add key of // left child if (isLeaf(root->left)) res += root->left->key; else { // Else recur for left child of root res += leftLeavesSum(root->left); } // Recur for right child of root and update res res += leftLeavesSum(root->right); } // return result return res;}This solution passes in a sum variable as an accumulator. When a left leaf is encountered, the leaf’s data is added to sum.
/* Pass in a sum variable as an accumulator */void leftLeavesSumRec(Node *root, bool isleft, int *sum){ if (!root) return; // Check whether this node is a leaf node and is left. if (root->left==null && root->right==null && isleft) *sum += root->key; // Pass 1 for left and 0 for right leftLeavesSumRec(root->left, 1, sum); leftLeavesSumRec(root->right, 0, sum);}