236. Lowest Common Ancestor of a Binary Tree

Difficulty: Medium

Related Topics: Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the : “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, 10].
• -10 <= Node.val <= 10
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

Solution

Language: Java

class Solution {

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // 出口
        // 1. 没有 LACA
        if (root == null) {
            return null;
        }
        // 2. 节点自己也可以是自己的祖先
        //
        //    LCA -> q          p <- LCA
        //          /            \
        //         p              q
        //
        if (root == p || root == q) {
            return root;
        }

        // 1. 往左子树搜索
        // 2. 往右子树搜索
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        // left == null 表示没有在左子树中找到 p 或 q
        // 因为题目说了 p 和 q 是在 Tree 上的点
        // 如果没有在左子树找到 p 或 q，那么右子树中必然有 p 或 q
        // 所以把我们在右子树中找到的结果返回去，充当局部解
        if (left == null) {
            return right;
        }

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

        // 如果我们能在左子树中找到 p 或 q 且在右子树中找到 p 或 q
        // 那么 root 就是 LCA
        return root;
    }
}