865. Smallest Subtree with all the Deepest Nodes

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865. Smallest Subtree with all the Deepest Nodes

Description

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 .

Note: This question is the same as 1123: https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/

Hints/Notes

  • binary tree

Solution

Language: C++

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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* res;
    int maxDepth = -1;

    TreeNode* subtreeWithAllDeepest(TreeNode* root) {
        if (!root) {
            return root;
        }

        traverse(root, 0);
        return res;
    }

    int traverse(TreeNode* root, int depth) {
        if (!root) {
            return depth - 1;
        }

        if (depth > maxDepth) {
            maxDepth = depth;
            res = root;
        }

        int left = traverse(root->left, depth + 1);
        int right = traverse(root->right, depth + 1);

        if (left == right && left == maxDepth) {
            res = root;
        }

        return left > right ? left : right;
    }
};