236 Lowest Common Ancestor of a Binary Tree.md

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

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

Solution

记录父节点法

/**
 * 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 {
public:
    unordered_map<int, TreeNode*> son_father;
    unordered_map<int, bool> visited;

    void dfs(TreeNode* root){
        if (root->left != nullptr) {
            son_father[root->left->val] = root;
            dfs(root->left);
        }
        if (root->right != nullptr) {
            son_father[root->right->val] = root;
            dfs(root->right);
        }
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        son_father[root->val] = nullptr;
        dfs(root);
        while (p != nullptr) {
            visited[p->val] = true;
            p = son_father[p->val];
        }
        while (q != nullptr) {
            if (visited[q->val]) return q;
            q = son_father[q->val];
        }
        return nullptr;
    }
};

执行用时：24 ms, 在所有 C++ 提交中击败了23.06%的用户

内存消耗：16.9 MB, 在所有 C++ 提交中击败了9.64%的用户

Solution

递归法

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

执行用时：8 ms, 在所有 C++ 提交中击败了99.50%的用户

内存消耗：14 MB, 在所有 C++ 提交中击败了32.04%的用户