865 Smallest Subtree with all the Deepest Nodes 具有所有最深节点的最小子树
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:
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
题目描述:
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。
注意：
本题与力扣 1123 重复：
示例 :
示例 1：
输入：root = [3,5,1,6,2,0,8,null,null,7,4]
输出：[2,7,4]
解释：
我们返回值为 2 的节点，在图中用黄色标记。
在图中用蓝色标记的是树的最深的节点。
注意，节点 5、3 和 2 包含树中最深的节点，但节点 2 的子树最小，因此我们返回它。
示例 2：
输入：root = [1]
输出：[1]
解释：根节点是树中最深的节点。
示例 3：
输入：root = [0,1,3,null,2]
输出：[2]
解释：树中最深的节点为 2 ，有效子树为节点 2、1 和 0 的子树，但节点 2 的子树最小。
提示:
树中节点的数量介于 1 和 500 之间。
0 <= Node.val <= 500
每个节点的值都是独一无二的。
思路:
DFS
这题目也太绕了, 题意是找到一棵树, 这棵树上所有的结点或者子结点包含深度最深的叶子结点
其实就是找到左子树和右子树高度相等的根结点
对每个结点查找高度即可
时间复杂度为 O(n), 空间复杂度为 O(n)
代码:
C++:
/**
* 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)
{
if (!root) return root;
int left = depth(root -> left), right = depth(root -> right);
return left > right ? subtreeWithAllDeepest(root -> left) : (left < right ? subtreeWithAllDeepest(root -> right) : root);
}
private:
int depth(TreeNode* root)
{
return !root ? 0 : max(depth(root -> left), depth(root -> right)) + 1;
}
};
Java:
/**
* 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) {
if (root == null) return root;
int left = depth(root.left), right = depth(root.right);
return left > right ? subtreeWithAllDeepest(root.left) : (left < right ? subtreeWithAllDeepest(root.right) : root);
}
private int depth(TreeNode root) {
return root == null ? 0 : Math.max(depth(root.left), depth(root.right)) + 1;
}
}
Python:
# 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: TreeNode) -> TreeNode:
def depth(node: TreeNode) -> int:
return 0 if not node else max(depth(node.left), depth(node.right)) + 1
return root if not root or (l := depth(root.left)) == (r := depth(root.right)) else self.subtreeWithAllDeepest(root.left) if l > r else self.subtreeWithAllDeepest(root.right)