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).”
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.
Example 3:
Input: root = [1,2], p = 1, q = 2
Output: 1
Constraints:
- The number of nodes in the tree is in the range [2, 105].
- -109 <= Node.val <= 109
- All Node.val are unique.
- p != q
- p and q will exist in the tree.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
// 根结点为空，直接返回
if(root == null){
return null;
}
// 根结点==p或者根结点==q，说明是个公共祖先
if(root.val==p.val || root.val==q.val){
return root;
}
// 左节点中找祖先
TreeNode lca1=lowestCommonAncestor(root.left,p,q);
// 右节点中找到祖先
TreeNode lca2=lowestCommonAncestor(root.right,p,q);
// 左右都不为空，说明根结点是
if(lca1 !=null && lca2 !=null){
return root;
}
// 左节点不为空，左节点是
if(lca1 !=null){
return lca1;
// 右节点不为空，右节点是
}else{
return lca2;
}
}
}