力扣 236. 二叉树的最近公共祖先

🔗 leetcode.cn/problems/lo…

题目

• 找到二叉树中，两个节点的公共祖先

思路

• 遍历二叉树，记录节点的 height 和 father
• 先让 p 和 q 的 height 保持一致，随后同步找 father，直到碰到相等的节点，即为最近的公共祖先

代码

/**
 * 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<TreeNode*, TreeNode*> fa_mp;
    unordered_map<TreeNode*, int> height_mp;
    void build(TreeNode* node) {
        if (node == nullptr) return;
        int height = height_mp[node];
        if (node->left) {
            height_mp[node->left] = height + 1;
            fa_mp[node->left] = node;
            build(node->left);
        }
        if (node->right) {
            height_mp[node->right] = height + 1;
            fa_mp[node->right] = node;
            build(node->right);
        }
        
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        fa_mp[root] = nullptr;
        height_mp[root] = 0;
        build(root);
        while (height_mp[p] < height_mp[q]) {
            q = fa_mp[q];
        }

        while (height_mp[p] > height_mp[q]) {
            p = fa_mp[p];
        }

        
        while (p != q) {
            p = fa_mp[p];
            q = fa_mp[q];
        }

        return p;
    }
};