896 Smallest Subtree With All The Deepest Nodes
896 Smallest Subtree With All The Deepest Nodes
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 a tree consisting of that node, plus the set of all descendants of that node.
Example 1:
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**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:
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**Input:** root = [1]
**Output:** [1]
**Explanation:** The root is the deepest node in the tree.
Example 3:
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**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:
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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/
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# 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]:
# lvl ord trav
# mainntain each levl in array
# latest lvl with count 1
r = []
arr = [root]
def recurse(arr, r):
if len(arr) == 0:
return
temp = []
r.append(copy.deepcopy(arr))
while(arr):
n = arr.pop()
if n.left:
temp.append(n.left)
if n.right:
temp.append(n.right)
recurse(temp, r)
recurse(arr, r)
#print(r[-1])
r = r[-1]
# r has leaf nodes
ans = {"ans": None}
def is_valid(root, r, count):
if root == None:
return
#print(root, count)
if root.val in map(lambda x: x.val, r):
count["c"] += 1
return
is_valid(root.left, r, count)
is_valid(root.right, r, count)
def dfs(root, r):
if not root:
return
count = {"c": 0}
is_valid(root, r , count)
#print(count, r)
if count["c"] == len(r):
ans["ans"] = root
dfs(root.left, r)
dfs(root.right, r) # only one of these will be correct ,
# if both are true that means parent is the ans
dfs(root, r)
return ans["ans"]
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