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]
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.
题目链接：https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/
思路
两种方法都可以用递归和非递归实现，这里只做了递归方法。
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法一：找路径，再找两个链的最低共同点
先找到两个节点的路径，然后两条路径都从根节点开始，向下找第一个分叉点。
/**
* 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 || !p || !q) return NULL;
auto pp = path(root, p);
auto pq = path(root, q);
TreeNode* res;
while(!pp.empty() && !pq.empty() && pp.top() == pq.top()){
res = pp.top();
pp.pop();
pq.pop();
}
return res;
}
stack<TreeNode*> path(TreeNode* root, TreeNode* p ){
stack<TreeNode*> res;
if(!root) return res;
if(root==p){
res.push(root);
return res;
}else{
auto l = path(root->left, p);
auto r = path(root->right, p);
if(l.size()>0){
l.push(root);
return l;
}
else if(r.size()>0){
r.push(root);
return r;
}
else return res;
}
}
};法二：标帜
找第一个满足子树上有两个节点的根节点。
/**
* 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* res = NULL;
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(!root || !p || !q) return NULL;
find(root, p, q);
return res;
}
bool find(TreeNode* root, TreeNode* p, TreeNode* q){
if(!root) return NULL;
int l = (find(root->left, p, q))?1:0;
int r = (find(root->right, p, q))?1:0;
int m = (root==p || root==q)?1:0;
cout<<root->val<<":"<<l<<","<<r<<","<<m<<endl;
if(l+r+m>=2){
res = root;
cout<<root->val;
}
return (l+r+m)>0;
}
};