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。
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 }
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