[Leetcode] 236. Lowest Common Ancestor of a Binary Tree

No236. The nearest common ancestor of a binary tree

topic

Given a binary tree, find the nearest common ancestor of two specified nodes in the tree.

The definition of the nearest common ancestor in Baidu Encyclopedia is: "For the two nodes p and q of the rooted tree T, the nearest common ancestor is expressed as a node x, so that x is the ancestor of p and q and the depth of x is as large as possible (A node can also be its own ancestor)."

For example, given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1

• Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
• Output: 3
• Explanation: The nearest common ancestor of node 5 and node 1 is node 3.

Example 2

• Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
• Output: 5
• Explanation: The nearest common ancestor of node 5 and node 4 is node 5. Because by definition the nearest common ancestor node can be the node itself.

Description:

• The values ​​of all nodes are unique.
• p and q are different nodes and both exist in the given binary tree.

Ideas

Using recursive thinking, consider the following operations:

• If the current node is empty and the value is equal to the value of p or q, it returns root;
• Otherwise, the left and right subtrees are recursively called to determine whether there is a result, p, q, and root have the following three relationships:
• [0,2], [1,1], [2,0], considering the first case, left must be empty, just return right; considering the third case, right must be empty, just return left; In the second case, both left and right are non-empty, just return to root directly.

Problem-solving code (Python3)

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if not root or root.val in [p.val,q.val]:
            return root 
        left = self.lowestCommonAncestor(root.left,p,q)
        right = self.lowestCommonAncestor(root.right,p,q)
        if not left:
            return right
        if not right:
            return left
        return root

Complexity analysis:

• The time complexity is O(n). The worst is to traverse all the nodes and drop the constant.
• Space complexity O(n) requires additional space to store the stack

operation result