LeetCode: 236. Lowest Common Ancestor of a Binary Tree

题目

原题地址：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia : “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Example 1:

image1

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

image2

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints:

  • The number of nodes in the tree is in the range [2, 10^5].
  • -10^9 <= Node.val <= 10^9
  • All Node.val are unique.
  • p != q
  • p and q will exist in the tree.

题目大意是，求二叉树中指定两个节点的最近共同祖先。

解法

遍历二叉树：

  • 如果当前 root 节点为 None ，则返回 None
  • 如果当前 root 节点节点值等于 p 或 q 的值，则当前节点即为要找的 LCA，因为当前节点是 p 或 q 其中一个节点，不会有比它更近的共同祖先了。
  • 在左子树中查找，假设结果为 left
  • 在右子树中查找，假设结果为 right
  • 如果 left 和 right 都不为 None，说明 left 和 right 刚好就是 p 和 q 这两个节点， 那么当前 root 节点即为要找的 LCA
  • 否则的话，left 和 right 中哪个不为 None，哪个就是要找的 LCA

这个思路的 Python 代码类似下面这样：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        if root is None:
            return None

        if root.val == p.val or root.val == q.val:
            return root

        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)

        if left is not None and right is not None:
            return root

        if left is not None:
            return left
        if right is not None:
            return right

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