题目:
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 nodes5
and1
is3.
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 nodes5
and4
is5
, 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.
实现:
- 递归方式,从根节点开始查找 p和q,如果查找到直接返回root,不是从左右子树分配查找, 如果左子树为空,那么找右子树, 如果右子树为空找左子树。
- 时间复杂度 O(N)
/**C++
* 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 || root == q || root == NULL){
return root;
}
TreeNode *left = lowestCommonAncestor(root->left, p, q);
TreeNode *right= lowestCommonAncestor(root->right, p, q);
//return left == NULL? right : right == NULL ? left: root;
if (left == NULL )
return right;
if (right == NULL)
return left;
return root;
}
};
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