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).

 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. 

方法

深度优先遍历法（DFS）

• 时间复杂度：O(n)
• 空间复杂度：O(n)

class Solution {
    private TreeNode result = null;
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        dfs(root,p,q);
        return result;
    }
    private boolean dfs(TreeNode root, TreeNode p, TreeNode q){
        if(root==null){
            return false;
        }
        boolean left = dfs(root.left,p,q);
        boolean right = dfs(root.right,p,q);
        if(left&&right || ((left||right)&&(root.val==p.val||root.val==q.val))) {
            result = root;
        }
        return left || right || (root.val==p.val || root.val==q.val);
    }
}

存储父节点法

• 时间复杂度：O(n)
• 空间复杂度：O(n)