Smallest Subtree With All The Deepest Nodes Problem
Description
LeetCode Problem 865.
Given the root of a binary tree, the depth of each node is the shortest distance to the root.
Return the smallest subtree such that it contains all the deepest nodes in the original tree. A node is called the deepest if it has the largest depth possible among any node in the entire tree.
The subtree of a node is a tree consisting of that node, plus the set of all descendants of that node.
Example 1:
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Input: root = [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 coloured in blue are the deepest nodes of the tree.
Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2 is the smallest subtree among them, so we return it.
Example 2:
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Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree.
Example 3:
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Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest node in the tree is 2, the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.
Constraints:
- The number of nodes in the tree will be in the range [1, 500].
- 0 <= Node.val <= 500
- The values of the nodes in the tree are unique.
Sample C++ Code
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/**
* 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:
TreeNode* subtreeWithAllDeepest(TreeNode* root) {
if (root == NULL)
return NULL;
int left = getDepth(root->left);
int right = getDepth(root->right);
if (left == right)
return root;
else if (left > right)
return subtreeWithAllDeepest(root->left);
else return
subtreeWithAllDeepest(root->right);
}
private :
int getDepth(TreeNode* node) {
if (node == NULL)
return 0;
return 1 + std::max(getDepth(node->left), getDepth(node->right));
}
};