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Find all leaf node in a binary tree

What are leaf nodes? 

Leaf nodes are those nodes in a tree which have no left and no right child. our problem is to find out all the leaf nodes in a binary tree and store them in a vector.

find out all leaf nodes in a binary tree
binary tree


In this figure , the leaf nodes are :

[8 , 9 , 10 , 11 , 13 , 14]



Suppose we have :

Struct TreeNode
{
Int val ;
TreeNode* left ;
TreeNode* right;
};

Form a new node : 

TreeNode* node = new TreeNode();              // c++ code

Or 

TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode)); // c code

After applying it to tree , we will check it the node is a leaf node.

To check whether this node is leaf node or not   : 


if(node->left == NULL && node->right == NULL)      // c and c++ code


If it return true , it means `node` is a leaf node else it will other `node`.


How to find all  leaf nodes in a binary tree?


Here we will find all the leaf nodes in a binary tree and store these in a vector. 

Algorithm : 

  • Use level order traversal to traverse all the nodes in a binary tree. 

  • For every node check the above condition if the node is leaf node or not.

  • If it is not a leaf node ,then move forward else push it in the vector.



Code : 

void leaf_nodes(TreeNode* root , vector<TreeNode*> &leafs)
    {
        queue<TreeNode*>q;
        q.push(root);
        while(!q.empty())
        { 
            TreeNode* node = q.front();
            if(node->left == NULL && node->right == NULL)
            {
                leafs.push_back(node);
            }
            if(node->left)
            {
                q.push(node->left);
            }
            if(node->right)
            {
                q.push(node->right);
            }
            q.pop();
        }

    }


References : 

https://mycodevillage.blogspot.com/2022/09/429-n-ary-tree-level-order-traversal.html


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