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, 104]. -231 <= Node.val <= 231 - 1
**Java Solution**
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
public int findLeftMostNode(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
root = queue.poll();
if (root.right != null)
queue.add(root.right);
if (root.left != null)
queue.add(root.left);
}
return root.val;
}
**C++ Solution**
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) {
vector<int> res;
bottomLeft(root, res,0);
return res[res.size()-1];
}
void bottomLeft(TreeNode * root, vector<int> &res, int h){
if(root==NULL)
return;
if(res.size()==h){
res.push_back(root->val);
}
bottomLeft(root->left, res,h+1);
bottomLeft(root->right, res,h+1);
}
};