LeetCode 1644. Lowest Common Ancestor of a Binary Tree II

Tree

Given the root of a binary tree, return the lowest common ancestor (LCA) of two given nodes, p and q. If either node p or q does not exist in the tree, return null. All values of the nodes in the tree are unique.

According to the definition of LCA on Wikipedia: "The lowest common ancestor of two nodes p and q in a binary tree T is the lowest node that has both p and q as descendants (where we allow a node to be a descendant of itself)". A descendant of a node x is a node y that is on the path from node x to some leaf node.

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. A node can be a descendant of itself according to the definition of LCA.

Example 3:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 10
Output: null
Explanation: Node 10 does not exist in the tree, so return null.

Constraints:

• The number of nodes in the tree is in the range [1, 104].

• -109 <= Node.val <= 109

• All Node.val are unique.

• p != q

Follow up: Can you find the LCA traversing the tree, without checking nodes existence?

Solution

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中文版解答Bilibili Link

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    
    pair<TreeNode*, int> dfs(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root == nullptr) {
            return {nullptr, 0};
        }
    
        pair<TreeNode*, int> l_res = dfs(root->left, p, q);
        pair<TreeNode*, int> r_res = dfs(root->right, p, q);
        
        if (root == p || root == q) {
            return {root, 1 + l_res.second + r_res.second};
        }
    
        if (l_res.first && r_res.first) {
            return {root, 2};
        }
    
        return l_res.first ? l_res : r_res;
    }
        
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        pair<TreeNode*, int> res = dfs(root, p, q);
        if (res.second < 2) {
            return nullptr;
        }
        return res.first;
    }
};