Tree - 236. Lowest Common Ancestor of a Binary Tree

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).”

Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]

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.

Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the binary tree.

思路：

这题和235区别在于给出的树是普通的二叉树，而不是bst，至于做法就是后序遍历，找出左子树是否是target或者右子树有没有target，如果左子树和右子树分别有一个target，那么LCA就是当前节点，如果只有一个目标命中，那么就把命中目标返回。

代码：

go：

/**

 * Definition for TreeNode.

 * type TreeNode struct {
 *     Val int
 *     Left *ListNode
 *     Right *ListNode
 * }
 */
 func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
     if root == nil || root == p || root == q {
         return root
     }
     
     left := lowestCommonAncestor(root.Left, p, q)
     right := lowestCommonAncestor(root.Right, p, q)
     
     if left != nil && right != nil {
         return root
     }
     
     if left == nil {
         return right
     } else {
         return left
     }
}