【LeetCode】236. Lowest Common Ancestor of a Binary Tree 解題報告

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236. Lowest Common Ancestor of a Binary Tree / Medium

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
utput: 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

Comstraints

  • 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.

Solution: Recursive

思路

我們先一樣尋訪整棵樹，如果遇到某個節點是 p 或是 q 時，就回傳，否則就是 nullptr。
如果某個節點收到左右兩邊回傳的都不是 nullptr，代表這個節點就是最小共同祖先了。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (!root) return nullptr;
        if (root == p || root == q) return root;
        
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        
        if (left && right) return root;
        
        return left ? left : right;
    }
};