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This approach involves using recursive DFS to traverse the tree. At each node, determine if it is a left leaf. If it is, add its value to the sum and recursively continue to search for more left leaves in the subtree.
The time complexity is O(n) where n is the number of nodes, as we visit each node once. The space complexity is O(h), where h is the height of the tree, due to the recursive call stack.
1class TreeNode {
2 int val;
3 TreeNode left;
4 TreeNode right;
5 TreeNode(int x) { val = x; }
6}
7
8public class Solution {
9 public int dfs(TreeNode node, boolean isLeft) {
10 if (node == null) return 0;
11 if (node.left == null && node.right == null && isLeft) return node.val;
12 return dfs(node.left, true) + dfs(node.right, false);
13 }
14 public int sumOfLeftLeaves(TreeNode root) {
15 return dfs(root, false);
16 }
17}In Java, we implement the DFS using a recursive method. The solution leverages the dfs method to accumulate the sum of values for left leaves by recursively checking each subtree.
This approach applies iterative DFS using a stack. By exploring each node iteratively, it checks if a node qualifies as a left leaf and accumulates its value to the sum. It manages to mimic a recursive pattern iteratively.
The time complexity is O(n) for navigating each tree node once. Space complexity is O(n) due to maintaining a stack proportional to the number of nodes.
1#include <stack>
using namespace std;
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
int sumOfLeftLeaves(TreeNode* root) {
if (!root) return 0;
int sum = 0;
stack<pair<TreeNode*, bool>> stk;
stk.push({root, false});
while (!stk.empty()) {
TreeNode* node = stk.top().first;
bool isLeft = stk.top().second;
stk.pop();
if (!node->left && !node->right && isLeft) {
sum += node->val;
}
if (node->right) {
stk.push({node->right, false});
}
if (node->left) {
stk.push({node->left, true});
}
}
return sum;
}
};The iterative C++ method represents nodes and their 'left-child' status using a stack. As the nodes traverse the binary tree, any left leaf node's value gets summed up to achieve the final result.