​LeetCode刷题实战236：二叉树的最近公共祖先

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        stack<TreeNode *> stack;
        stack.push(root);
        map<TreeNode *, TreeNode *> parent_map;
        parent_map[root] = NULL;
        while(parent_map.find(p) == parent_map.end() || parent_map.find(q) == parent_map.end()){
            TreeNode *node = stack.top();
            stack.pop();
            if(node -> left != NULL){
                parent_map[node -> left] = node;
                stack.push(node -> left);
            }
            if(node -> right != NULL){
                parent_map[node -> right] = node;
                stack.push(node -> right);
            }
        }
        set<TreeNode *> p_parent;
        p_parent.insert(p);
// 迭代父节点map，这里的循环逻辑为，父节点字典中如果一直能够找到p节点的父节点，就进入循环
// 这样可以一直得到p节点的父节点列表。
        while(parent_map.find(p) != parent_map.end()){
            p_parent.insert(parent_map[p]);
            p = parent_map[p];
        }
// 同理遍历q节点的父节点列表，因为是逆序遍历，所以当它存在于p节点的列表时，找到的就是p和q的最近公共祖先
        while(parent_map.find(q) != parent_map.end()){
            if(p_parent.find(q) != p_parent.end()){
                return q;
            }
            q = parent_map[q];
        }
        return root;
    }
};

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        int num = contain_node(root, p, q);
        return ans;
    }
    int contain_node(TreeNode* root, TreeNode* p, TreeNode* q){
        if(!root) return 0;
        int mid = 0;
        if(root == p || root == q) mid = 1;
        int left = contain_node(root -> left, p, q);
        if(mid + left == 2){
            if(!ans) ans = root;
            return 2;
        }
        int right = contain_node(root -> right, p, q);
        if(left + right + mid >= 2){
            if(!ans) ans = root;
            return 2;
        }
        return left + mid + right;
    }
private:
    TreeNode* ans = NULL;
};