# Find Bottom Left Tree Value | Breadth-First search

Gowthaman Ravi
·Feb 6, 2022·

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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) {
while (!queue.isEmpty()) {
root = queue.poll();
if (root.right != null)
if (root.left != null)
}
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);
}
};
``````