865. Smallest Subtree with all the Deepest Nodes

Given a binary tree rooted at root, the depth of each node is the shortest distance to the root.

A node is deepest if it has the largest depth possible among any node in the entire tree.

The subtree of a node is that node, plus the set of all descendants of that node.

Return the node with the largest depth such that it contains all the deepest nodes in its subtree.

Example 1:

Input: [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation:



We return the node with value 2, colored in yellow in the diagram.
The nodes colored in blue are the deepest nodes of the tree.
The input "[3, 5, 1, 6, 2, 0, 8, null, null, 7, 4]" is a serialization of the given tree.
The output "[2, 7, 4]" is a serialization of the subtree rooted at the node with value 2.
Both the input and output have TreeNode type.

Note:

• The number of nodes in the tree will be between 1 and 500.
• The values of each node are unique.

The thought process goes like this - Find all deepest nodes by traversing the tree using BFS. The required node is nothing but the LCA of all the deepest nodes. Finding LCA of all nodes at the same level is equivalent to finding LCA of the leftMost and rightMost node. Keep track of leftMost and rightMost nodes while doing BFS and finally return their LCA.

class Solution {
public:
    TreeNode* lca( TreeNode* root, TreeNode* p, TreeNode* q ) {
        if ( !root || root == p || root == q ) return root;
        TreeNode *left = lca( root->left, p, q );
        TreeNode *right = lca (root->right, p, q );

        return !left? right: !right? left: root;
    }
    
    TreeNode* subtreeWithAllDeepest(TreeNode* root) {
        if ( !root || !root->left && !root->right ) return root;
        TreeNode *leftMost = NULL;
        TreeNode *rightMost = NULL;
        
        queue<TreeNode*> q;
        q.push(root);
        while( !q.empty() ) {
            int levelSize = q.size();
            for(int level = 0; level < levelSize; level++ ) {
                TreeNode* node = q.front(); q.pop();
                if ( level == 0 ) leftMost = node;
                if ( level == levelSize - 1 ) rightMost = node;
                
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
                
            }
        }
        return lca( root, leftMost, rightMost );
    }
};