865. Smallest Subtree with all the Deepest Nodes ⭐


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.


  • 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/


 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
func subtreeWithAllDeepest(root *TreeNode) *TreeNode {
    diff := dfs(root.Left, 0) - dfs(root.Right, 0) //左右的深度差

    if diff == 0 { //左右最深的深度相同,代表當下這個節點是最小子樹,回傳
        return root

    if diff > 0 {  //左右深度不同,且最深的在左邊,從左子節點重新檢查是否左右深度相同
        return subtreeWithAllDeepest(root.Left) 

    return subtreeWithAllDeepest(root.Right)

func dfs(root *TreeNode, depth int) int {
    if root == nil { return depth }
    return max(dfs(root.Left, depth+1), dfs(root.Right, depth+1))

func max(a, b int) int {
    if a > b { return a }
    return b

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