[leetcode] 1123. Lowest Common Ancestor of Deepest Leaves

Description

Given the root of a binary tree, return the lowest common ancestor of its deepest leaves.

Recall that:

• The node of a binary tree is a leaf if and only if it has no children
• The depth of the root of the tree is 0. if the depth of a node is d, the depth of each of its children is d + 1.
• The lowest common ancestor of a set S of nodes, is the node A with the largest depth such that every node in S is in the subtree with root A.

Note: This question is the same as 865: https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/

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 leaf-nodes of the tree.
Note that nodes 6, 0, and 8 are also leaf nodes, but the depth of them is 2, but the depth of nodes 7 and 4 is 3.

Example 2:

Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree, and it's the lca of itself.

Example 3:

Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest leaf node in the tree is 2, the lca of one node is itself.

Constraints:

• The number of nodes in the tree will be in the range [1, 1000].
• 0 <= Node.val <= 1000
• The values of the nodes in the tree are unique.

分析

题目的意思是：找出最深的叶子结点的最小公共祖先，如果叶子结点只有一个，则公共祖先就是其自身。这道题递归的话需要记录深度和最小的祖先结点，进行后序遍历。如果左右子树返回的深度不相等，则最小祖先节点在更深的分支上。如果相等，则最小祖先节点就是当前的节点。

代码

# 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 helper(self,root):
        if(root is None):
            return (0,None)
        l,lnode=self.helper(root.left)
        r,rnode=self.helper(root.right)
        if(l<r):
            return r+1,rnode
        elif(l>r):
            return l+1,lnode
        return l+1,root
        
    def lcaDeepestLeaves(self, root: TreeNode) -> TreeNode:
        l,node=self.helper(root)
        return node

参考文献

[LeetCode]Python3 concise solution