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:

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:

Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree.

Example 3:

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/

Solutions

Solution 1: Recursion

We design a function $\textit{dfs}(\textit{root})$ that returns the smallest subtree containing all the deepest nodes in the subtree rooted at $\textit{root}$, as well as the depth of the subtree rooted at $\textit{root}$.

The execution process of the function $\textit{dfs}(\textit{root})$ is as follows:

• If $\textit{root}$ is null, return $\text{null}$ and $0$.
• Otherwise, recursively calculate the smallest subtree and depth of the left and right subtrees of $\textit{root}$, denoted as $l$ and $l_d$, and $r$ and $r_d$, respectively. If $l_d > r_d$, then the smallest subtree containing all the deepest nodes in the subtree rooted at the left child of $\textit{root}$ is $l$, with a depth of $l_d + 1$. If $l_d < r_d$, then the smallest subtree containing all the deepest nodes in the subtree rooted at the right child of $\textit{root}$ is $r$, with a depth of $r_d + 1$. If $l_d = r_d$, then $\textit{root}$ is the smallest subtree containing all the deepest nodes, with a depth of $l_d + 1$.

Finally, return the first element of the result of $\textit{dfs}(\textit{root})$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

Python3JavaC++GoTypeScript

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def subtreeWithAllDeepest(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def dfs(root: Optional[TreeNode]) -> Tuple[Optional[TreeNode], int]:
            if root is None:
                return None, 0
            l, ld = dfs(root.left)
            r, rd = dfs(root.right)
            if ld > rd:
                return l, ld + 1
            if ld < rd:
                return r, rd + 1
            return root, ld + 1

        return dfs(root)[0]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode subtreeWithAllDeepest(TreeNode root) {
        return dfs(root).getKey();
    }

    private Pair<TreeNode, Integer> dfs(TreeNode root) {
        if (root == null) {
            return new Pair<>(null, 0);
        }
        var l = dfs(root.left);
        var r = dfs(root.right);
        int ld = l.getValue(), rd = r.getValue();
        if (ld > rd) {
            return new Pair<>(l.getKey(), ld + 1);
        }
        if (ld < rd) {
            return new Pair<>(r.getKey(), rd + 1);
        }
        return new Pair<>(root, ld + 1);
    }
}

/**
 * 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) {
        using pti = pair<TreeNode*, int>;
        auto dfs = [&](auto&& dfs, TreeNode* root) -> pti {
            if (!root) {
                return {nullptr, 0};
            }
            auto [l, ld] = dfs(dfs, root->left);
            auto [r, rd] = dfs(dfs, root->right);
            if (ld > rd) {
                return {l, ld + 1};
            }
            if (ld < rd) {
                return {r, rd + 1};
            }
            return {root, ld + 1};
        };
        return dfs(dfs, root).first;
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func subtreeWithAllDeepest(root *TreeNode) *TreeNode {
    type pair struct {
        node  *TreeNode
        depth int
    }
    var dfs func(*TreeNode) pair