Question

865. Smallest Subtree with all the Deepest Nodes

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Description

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

 

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation: We return the node with value 2, colored in yellow in the diagram.
The nodes coloured in blue are the deepest nodes of the tree.
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:

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

Example 3:

Input: root = [0,1,3,null,2]
Output: [2]
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.

 

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

Solutions

Solution 1

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def subtreeWithAllDeepest(self, root: TreeNode) -> TreeNode:
    def dfs(root):
      if root is None:
        return None, 0
      l, d1 = dfs(root.left)
      r, d2 = dfs(root.right)
      if d1 > d2:
        return l, d1 + 1
      if d1 < d2:
        return r, d2 + 1
      return root, d1 + 1

    return dfs(root)[0]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  public TreeNode subtreeWithAllDeepest(TreeNode root) {
    return dfs(root).getKey();
  }

  private Pair<TreeNode, Integer> dfs(TreeNode root) {
    if (root == null) {
      return new Pair<>(null, 0);
    }
    Pair<TreeNode, Integer> l = dfs(root.left);
    Pair<TreeNode, Integer> r = dfs(root.right);
    int d1 = l.getValue(), d2 = r.getValue();
    if (d1 > d2) {
      return new Pair<>(l.getKey(), d1 + 1);
    }
    if (d1 < d2) {
      return new Pair<>(r.getKey(), d2 + 1);
    }
    return new Pair<>(root, d1 + 1);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
using pti = pair<TreeNode*, int>;
class Solution {
public:
  TreeNode* subtreeWithAllDeepest(TreeNode* root) {
    return dfs(root).first;
  }

  pti dfs(TreeNode* root) {
    if (!root) return {nullptr, 0};
    pti l = dfs(root->left);
    pti r = dfs(root->right);
    int d1 = l.second, d2 = r.second;
    if (d1 > d2) return {l.first, d1 + 1};
    if (d1 < d2) return {r.first, d2 + 1};
    return {root, d1 + 1};
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
type pair struct {
  first  *TreeNode
  second int
}

func subtreeWithAllDeepest(root *TreeNode) *TreeNode {
  var dfs func(root *TreeNode) pair
  dfs = func(root *TreeNode) pair {
    if root == nil {
      return pair{nil, 0}
    }
    l, r := dfs(root.Left), dfs(root.Right)
    d1, d2 := l.second, r.second
    if d1 > d2 {
      return pair{l.first, d1 + 1}
    }
    if d1 < d2 {
      return pair{r.first, d2 + 1}
    }
    return pair{root, d1 + 1}
  }
  return dfs(root).first
}