题目

给定二叉树和两个点，求两点的LCA最近公共祖先

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

 

 

思路

是一道套路题，A B俩点的位置只有三种情况

- A B 都在root左边（这时LCA必也在左边）

- A B 都在root右边（这时LCA必也在右边）

- A B 分别在root的左右边（这时LCA必为root）

 

 

代码

写起来很简单，但没想通为什么回溯之后得到的就是LCA，怎么证明

 

class Solution {
        public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
            if(root==p||root==q||root==null)
                return root;
            TreeNode leftRoot=lowestCommonAncestor(root.left,p,q);
            TreeNode rightRoot=lowestCommonAncestor(root.right,p,q);

            if (leftRoot==null)
                return rightRoot;//root的left没有pq，即pq都在右边，LCA也在右边
            else if(rightRoot==null)
                return leftRoot;//root的right没有pq，即pq都在左边，LCA也在左
            else
                return root;//俩节点分别在root的左右俩边，那LCA就是rott本身
        }
}