树的两个节点的最低公共祖先

Lowest Common Ancestor of a Binary Tree

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

方法一
递归

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

方法二
深度搜索

class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        vector<TreeNode*> pp, pq;        dfs(root, p, pp);        dfs(root, q, pq);        TreeNode* result = nullptr;        int n = (int)std::min(pp.size(), pq.size());        for (int i = 0; i < n; i++){            if (pp[i] == pq[i])                result = pp[i];            else                break;        }        return result;    }    bool dfs(TreeNode* root, TreeNode* p, vector<TreeNode*>& pp) {        if (!root)             return false;        pp.push_back(root);        if (root == p || dfs(root->left, p, pp) || dfs(root->right, p, pp))            return true;        pp.pop_back();        return false;    }};