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236. Lowest Common Ancestor of a Binary Tree ​

题目 ​

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:

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:

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, 105].
  • -109 <= Node.val <= 109
  • All Node.val are unique.
  • p != q
  • p and q will exist in the tree.
Related Topics
  • 树
  • 深度优先搜索
  • 二叉树

  • 👍 3008
  • 👎 0

    • 思路 ​

    递归的三步曲

    • 函数的定义 参数以及返回值
      • 参数 当前节点
      • 返回值 最小公共父节点
    • 结束条件
      • 当前节点为目标节点时，即可返回
      • 当前节点是空节点
    • 单层逻辑
      • 两个节点关系，只有两种
      • 在不同子树上
      • left and right 同时存在
      • 在同一颗子树上
        • left or right 自始至终仅有一个为 true

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        res = root

        def dfs(root: 'TreeNode'):
            if not root or root == p or root == q:
                return root

            left = dfs(root.left)
            right = dfs(root.right)

            if left and right:
                return root

            return left or right

        res = dfs(root)

        return res



# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(H)