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;
    }
};

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