题目描述

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

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

 

Note:

• All of the nodes' values will be unique.

• p and q are different and both values will exist in the binary tree.

难度系数 Medium

解法一：简单的dfs

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (!root || root==p || root==q) return root;
        //得到的是子节点
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        //返回父节点
        if (left && right) return root;
        
        return left ? left : right;       
    }
};

 

解法二：对上个算法的优化

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
       if (!root || p == root || q == root) return root;
       TreeNode *left = lowestCommonAncestor(root->left, p, q);
       //表示已找到节点，不需要再找
       if (left && left != p && left != q) return left;
       TreeNode *right = lowestCommonAncestor(root->right, p , q);
　　　　if (left && right) return root;
       return left ? left : right;
    }
};

github地址：https://github.com/AntonioSu/leetcode/blob/master/problems/236.LowestCommonAncestorofaBinaryTree.md