236. Lowest Common Ancestor of a Binary Tree

O(n) 分治

/**
 * 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:
    // 在root为根的二叉树中找A,B的LCA:
    // 如果找到了就返回这个LCA
    // 如果只碰到A，就返回A
    // 如果只碰到B，就返回B
    // 如果都没有，就返回null
		// 这里默认树上一定有p和q，所以不一定要真的把p和q都找到，假设只找到了其中一个，没找到另一个，说明找到的这个就是lca，另一个一定在以找到的这个点为根的树上
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (!root || root == p || root == q) return root;
        auto l = lowestCommonAncestor(root->left, p, q);
        auto r = lowestCommonAncestor(root->right, p, q);
        if (l && r) return root;
        return l ? l : r;
    }
};

/**
 * 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:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        TreeNode *res;
        lca(root, p, q, res);
        return res;
    }

    void lca(TreeNode* root, TreeNode* p, TreeNode* q, TreeNode *&res) {
        if (!root || root == p || root == q) {
            res = root;
            return;
        }
        TreeNode *l, *r;
        lca(root->left, p, q, l);
        lca(root->right, p, q, r);
        if (!l) {
            res = r;
        } else if (!r) {
            res = l;
        } else {
            res = root;
        }
    }
};