[Leetcode]236.|[Leetcode]236. Lowest Common Ancestor of a Binary Tree
236. 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 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]
![[Leetcode]236.|[Leetcode]236. Lowest Common Ancestor of a Binary Tree](https://img.it610.com/image/info10/271b4308759f446aa6d7b0e74cca5b6c.jpg)
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二叉树 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.
题意：给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。
【[Leetcode]236.|[Leetcode]236. Lowest Common Ancestor of a Binary Tree】思路：
如果以root为根的子树中包含p和q，则返回它们的最近公共祖先
如果只包含p，则返回p
如果只包含q，则返回q
如果都不包含，则返回NULL
右子树情况：
1.右边也都不包含，则right=NULL，最终需要返回NULL
2.右边只包含p或q，则right=p或者q，最终需要返回p或q
3.右边同时包含p和q，则right是最近公共祖先，我们最终也需要返回最近公共祖先
/**
* 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) {
if(root==NULL||root==p||root==q)
return root;
TreeNode *left=lowestCommonAncestor(root->left,p,q);
TreeNode *right=lowestCommonAncestor(root->right,p,q);
if(left!=NULL&&right!=NULL)
return root;
//如果p,q刚好在左右两个子树上
if(left==NULL)
return right;
//仅在右子树
if(right==NULL)
return left;
//仅在左子树
return root;
}
};
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