Find Bottom Left Tree Value
Link
Description
Given the root
of a binary tree, return the leftmost value in the last row of the tree.
Example 1:
- Input:
root = [2,1,3]
- Output:
1
Example 2:
- Input:
root = [1,2,3,4,null,5,6,null,null,7]
- Output:
7
Constraints:
- The number of nodes in the tree is in the range
[1, 10^4]
.
-2^31 <= Node.val <= 2^31 - 1
Solution
Recursive Solution
| /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
private:
int res = 0;
int maxDepth = 0;
void traversal(TreeNode* root, int depth) {
if (!root) return;
if (depth > maxDepth) {
maxDepth = depth;
res = root->val;
}
traversal(root->left, depth + 1);
traversal(root->right, depth + 1);
}
public:
int findBottomLeftValue(TreeNode* root) {
traversal(root, 1);
return res;
}
};
|
Iterative Solution (Level-order)
| /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int findBottomLeftValue(TreeNode* root) {
int res;
queue<TreeNode*> q;
if (root) q.push(root);
while (!q.empty()) {
int len = q.size();
for (int i = 0; i < len; i++) {
auto t = q.front();
q.pop();
if (i == 0) res = t->val;
if (t->left) q.push(t->left);
if (t->right) q.push(t->right);
}
}
return res;
}
};
|