LeetCode: 236. Lowest Common Ancestor of a Binary Tree
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: 1root = [3,5,1,6,2,0,8,null,null,7,4]1
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.
解题思路 —— 递归求解
根据左右子树含有待求公共祖先节点的节点数,判断其是否为公共祖先节点。
AC 代码
/**
* 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
{
private:
// nodes:待求公共祖先节点的子节点
// nodesInTree: root 子树中包含的 nodes 中的节点的数目
TreeNode* lowestCommonAncestor(TreeNode* root, const vector<TreeNode*>& nodes, int&& nodesInTree)
{
int nodesInTreeRef;
return lowestCommonAncestor(root, nodes, nodesInTreeRef);
}
TreeNode* lowestCommonAncestor(TreeNode* root, const vector<TreeNode*>& nodes, int& nodesInTree)
{
if(root == nullptr)
{
return nullptr;
}
TreeNode* commonAncestor = nullptr;
// LCA 在 root 的左子树
int nodesInLeftTree = 0;
commonAncestor = lowestCommonAncestor(root->left, nodes, nodesInLeftTree);
if(nodesInLeftTree == nodes.size())
{
nodesInTree = nodesInLeftTree;
return commonAncestor;
}
// LCA 在 root 的右子树
int nodesInRightTree = 0;
commonAncestor = lowestCommonAncestor(root->right, nodes, nodesInRightTree);
if(nodesInRightTree == nodes.size())
{
nodesInTree = nodesInRightTree;
return commonAncestor;
}
// root 为 LCA
nodesInTree = nodesInLeftTree+nodesInRightTree;
if(find(nodes.begin(), nodes.end(), root) != nodes.end())
{
++nodesInTree;
}
if(nodesInTree == nodes.size()) return root;
return nullptr;
}
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
{
return lowestCommonAncestor(root, {p, q}, 0);
}
};