- 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.
解 对于root节点，如果左右子树分别有一个节点，则root是一个公共祖先
解1 递归
class Solution {
public:
TreeNode *ans;
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
LCA(root, p, q);
return ans;
}
bool LCA(TreeNode *root, TreeNode *p, TreeNode *q){
if(root == NULL)return false;
int l = LCA(root->left, p, q) ? 1 : 0; // p或者q在左子树
int r = LCA(root->right, p, q) ? 1 : 0; // p或者q在右子树
int mid = (root == p || root == q) ? 1 : 0; //找到了p或者q
if(mid + l + r >= 2)ans = root; // =2时，左右各一个；>2时，一个是另一个的先驱节点
return mid + l + r > 0; // >0表明找到了p或者q
}
};