题目

Given the root of a binary tree, the depth of each node is the shortest distance to the root.

Return the smallest subtree such that it contains all the deepest nodes in the original tree.

A node is called the deepest if it has the largest depth possible among any node in the entire tree.

The subtree of a node is tree consisting of that node, plus the set of all descendants of that node.

Note: This question is the same as 1123: https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/

Example 1:

image.png

  1. Input: root = [3,5,1,6,2,0,8,null,null,7,4]
  2. Output: [2,7,4]
  3. Explanation: We return the node with value 2, colored in yellow in the diagram.
  4. The nodes coloured in blue are the deepest nodes of the tree.
  5. Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2 is the smallest subtree among them, so we return it.

Example 2:

  1. Input: root = [1]
  2. Output: [1]
  3. Explanation: The root is the deepest node in the tree.

Example 3:

  1. Input: root = [0,1,3,null,2]
  2. Output: [2]
  3. Explanation: The deepest node in the tree is 2, the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.

Constraints:

  • The number of nodes in the tree will be in the range [1, 500].
  • 0 <= Node.val <= 500
  • The values of the nodes in the tree are unique.

题意

在二叉树中找到一个子树，使它包含所有深度最大的结点。

思路

先一次DFS找到所有最深的结点，然后用找公共祖先的方法递归处理：在当前结点的左子树和右子树中找包含最深结点的子树leftTree和rightTree，如果都存在，说明当前结点是一个公共祖先，返回该结点；如果只有leftTree存在，返回leftTree；如果只有rightTree存在，返回rightTree；如果都不存在，说明以当前结点为根结点的子树不在范围内，返回null。

代码实现

Java

  1. class Solution {
  2.     private Map<TreeNode, Integer> depth = new HashMap<>();
  3.     private int maxDepth = -1;
  5.     public TreeNode subtreeWithAllDeepest(TreeNode root) {
  6.         findDepth(root, 1);
  7.         return findAncestor(root);
  8.     }
  10.     private void findDepth(TreeNode node, int level) {
  11.         if (node == null) {
  12.             return;
  13.         }
  15.         maxDepth = Math.max(maxDepth, level);
  16.         depth.put(node, level);
  18.         findDepth(node.left, level + 1);
  19.         findDepth(node.right, level + 1);
  20.     }
  22.     private TreeNode findAncestor(TreeNode node) {
  23.         if (node == null || depth.get(node) == maxDepth) {
  24.             return node;
  25.         }
  27.         TreeNode leftAncestor = findAncestor(node.left);
  28.         TreeNode rightAncestor = findAncestor(node.right);
  30.         if (leftAncestor != null && rightAncestor != null) {
  31.             return node;
  32.         } else if (leftAncestor != null) {
  33.             return leftAncestor;
  34.         } else if (rightAncestor != null) {
  35.             return rightAncestor;
  36.         } else {
  37.             return null;
  38.         }
  39.     }
  40. }