Leetcode 865. Smallest Subtree with all the Deepest Nodes

题目：一棵二叉树有它的最大深度，找出包含了树中所有深度最大的结点的最深的子树。

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.

Difficulty:Medium

Category:

Analyze

这个题的模型其实比较左右子树的高度，如果左右子树的高度相等，说明当前节点就是要求的。这个解释是这样的：必须包含所有的最大高度的叶子，左右叶子高度相等，所以必须包含当前节点。

当左子树高度>右子树高度的时候，要求的节点在左边；反之，在右边。

所以，递归思路 + 一个pair。这个pair的思路是，保存了当前节点的深度和当前节点的最深子树节点。

Cite: LeetCode 865. Smallest Subtree with all the Deepest Nodes

Solution

Solution 1: Recursion

Time complexity: O(n) Space complexity: O(n)

Runtime: 12 ms, faster than 74.90% of C++ online submissions for Smallest Subtree with all the Deepest Nodes. Memory Usage: 15.3 MB, less than 44.90% of C++ online submissions for Smallest Subtree with all the Deepest Nodes.

class Solution {
 public:
  TreeNode* subtreeWithAllDeepest(TreeNode* root) { 
    return depth(root).second;
  }

 private:
  pair<int, TreeNode*> depth(TreeNode* root) {
    if (root == nullptr) return {-1, nullptr};
    auto left = depth(root->left);
    auto right = depth(root->right);
    int dl = left.first;
    int dr = right.first;
    return {max(dl, dr) + 1, dl == dr ? root : (dl > dr) ? left.second : right.second};
  }
};