LeetCode 1123. Lowest Common Ancestor of Deepest Leaves

原题链接在这里：https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/

题目：

Given a rooted binary tree, return the lowest common ancestor of its deepest leaves.

Recall that:

• The node of a binary tree is a leaf if and only if it has no children
• The depth of the root of the tree is 0, and if the depth of a node is d, the depth of each of its children is d+1.
• The lowest common ancestor of a set S of nodes is the node A with the largest depth such that every node in S is in the subtree with root A.

Example 1:

Input: root = [1,2,3]
Output: [1,2,3]
Explanation: 
The deepest leaves are the nodes with values 2 and 3.
The lowest common ancestor of these leaves is the node with value 1.
The answer returned is a TreeNode object (not an array) with serialization "[1,2,3]".

Example 2:

Input: root = [1,2,3,4]
Output: [4]

Example 3:

Input: root = [1,2,3,4,5]
Output: [2,4,5]

Constraints:

• The given tree will have between 1 and 1000 nodes.
• Each node of the tree will have a distinct value between 1 and 1000.

题解：

For dfs, state needs current TreeNode root only. return value needs both depth and lca of deepest leaves node.

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Divide and conquer, have left result and right result first. Then compare the depth, pick the bigger side, if both sides have same depth, return root with depth + 1.

Time Complexity: O(n). n is nodes count of the whole tree.

Space: O(logn).

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public TreeNode lcaDeepestLeaves(TreeNode root) {
12         if(root == null){
13             return root;
14         }
15         
16         return lcaDfs(root).node;
17     }
18     
19     private Pair lcaDfs(TreeNode root){
20         if(root == null){
21             return new Pair(root, 0);
22         }
23         
24         Pair l = lcaDfs(root.left);
25         Pair r = lcaDfs(root.right);
26         if(l.depth > r.depth){
27             return new Pair(l.node, l.depth + 1);
28         }else if(l.depth < r.depth){
29             return new Pair(r.node, r.depth + 1);
30         }else{
31             return new Pair(root, l.depth + 1);
32         }
33     }
34 }
35 
36 class Pair{
37     TreeNode node;
38     int depth;
39     public Pair(TreeNode node, int depth){
40         this.node = node;
41         this.depth = depth;
42     }
43 }