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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:

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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:

img

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

Solution:

分类讨论:

  1. 当前节点是空节点 \

  2. 当前节点是p --- 返回当前节点

  3. 当前节点是q /

  1. 其他: 是否找到p或q:

4.1 左右子树都找到 -- 返回当前节点

4.2 只有左子树找到 -- 返回递归左子树的结果

4.3 只有右子树找到 -- 返回递归右子树的结果

4.4 左右子数都没有找到 -- 返回空节点

/**
 * 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) {
       // base case 
       if (root == null){
           return null;
       } 

       if (root == p || root == q){
           return root;
       }

       TreeNode left = lowestCommonAncestor(root.left, p, q);
       TreeNode right = lowestCommonAncestor(root.right, p, q);

       if (left != null && right != null){
           return root;
       }

        if (left == null && right == null){
            return null;
        }

        if (left != null){
            return left;
        }else{
            return right;
        }

    }
}

// TC: O(n)
// SC: O(n)