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[LeetCode] 865. Smallest Subtree with all the Deepest Nodes

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 tree consisting of that node, plus the set of all descendants of that node.

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

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.

具有所有最深节点的最小子树。

给定一个根为 root 的二叉树，每个节点的深度是该节点到根的最短距离。

如果一个节点在整个树的任意节点之间具有最大的深度，则该节点是最深的。

一个节点的子树是该节点加上它的所有后代的集合。

返回能满足以该节点为根的子树中包含所有最深的节点这一条件的具有最大深度的节点。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/smallest-subtree-with-all-the-deepest-nodes
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这道题的思路类似236题，思路是DFS + memorization，这里我先提供一个遍历两次的思路。既然找的是一棵最小的子树，但是这棵子树包含一个深度最大的节点（也就是距离根节点最远的节点），那么首先我们要做的就是记录下树中每个节点的深度，这里我选择用hashmap记录。第二次遍历的时候我们用dfs分别去看左子树和右子树，看哪个子树上返回的深度更大，则递归去看那个子树。

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public TreeNode subtreeWithAllDeepest(TreeNode root) {
18         // corner case
19         if (root == null) {
20             return null;
21         }
22         // normal case
23         HashMap<TreeNode, Integer> map = new HashMap<>();
24         depth(root, map);
25         return dfs(root, map);
26     }
27 
28     private int depth(TreeNode root, HashMap<TreeNode, Integer> map) {
29         if (root == null) {
30             return 0;
31         }
32         if (map.containsKey(root)) {
33             return map.get(root);
34         }
35         int max = Math.max(depth(root.left, map), depth(root.right, map)) + 1;
36         map.put(root, max);
37         return max;
38     }
39 
40     private TreeNode dfs(TreeNode root, HashMap<TreeNode, Integer> map) {
41         int left = depth(root.left, map);
42         int right = depth(root.right, map);
43         if (left == right) {
44             return root;
45         }
46         if (left > right) {
47             return dfs(root.left, map);
48         } else {
49             return dfs(root.right, map);
50         }
51     }
52 }