Formatted question description: https://leetcode.ca/all/865.html

865. Smallest Subtree with all the Deepest Nodes (Medium)

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:

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.

Related Topics:
Tree, Depth-first Search, Breadth-first Search, Recursion

Solution 1.

## Solution 1.

```cpp
// OJ: https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/
// Time: O(N)
// Space: O(H)
class Solution {
    int maxDepth = -1, target = 0, cnt = 0;
    void count(TreeNode *root, int d) {
        if (!root) return;
        if (d > maxDepth) {
            target = 1;
            maxDepth = d;
        } else if (d == maxDepth) ++target;
        count(root->left, d + 1);
        count(root->right, d + 1);
    }
    TreeNode *find(TreeNode *root, int d) {
        if (!root) return NULL;
        int before = cnt;
        auto left = find(root->left, d + 1);
        if (left) return left;
        auto right = find(root->right, d + 1);
        if (right) return right;
        if (d == maxDepth) ++cnt;
        return before == 0 && cnt == target ? root : NULL;
    }
public:
    TreeNode* lcaDeepestLeaves(TreeNode* root) {
        count(root, 0);
        return find(root, 0);
    }
};

Solution 2.

The lowest ancester is the highest node whose left and right subtrees have the same height.

// OJ: https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/
// Time: O(N)
// Space: O(H)
class Solution {
    pair<TreeNode*, int> dfs(TreeNode *root, int d = 0) { // latest node which has equal depth in left and right sub-trees; the corresponding height
        if (!root) return {NULL, 0};
        const auto &[left, ld] = dfs(root->left, d + 1);
        const auto &[right, rd] = dfs(root->right, d + 1);
        if (ld > rd) return {left, ld + 1};
        else if (ld < rd) return{right, rd + 1};
        return {root, ld + 1};
    }
public:
    TreeNode* lcaDeepestLeaves(TreeNode* root) {
        return dfs(root).first;
    }
};

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode subtreeWithAllDeepest(TreeNode root) {
            Map<Integer, List<TreeNode>> depthNodesMap = new HashMap<Integer, List<TreeNode>>();
            Map<TreeNode, TreeNode> childParentMap = new HashMap<TreeNode, TreeNode>();
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.offer(root);
            int depth = 0;
            while (!queue.isEmpty()) {
                depth++;
                List<TreeNode> list = new ArrayList<TreeNode>();
                int size = queue.size();
                for (int i = 0; i < size; i++) {
                    TreeNode node = queue.poll();
                    list.add(node);
                    TreeNode left = node.left, right = node.right;
                    if (left != null) {
                        childParentMap.put(left, node);
                        queue.offer(left);
                    }
                    if (right != null) {
                        childParentMap.put(right, node);
                        queue.offer(right);
                    }
                }
                depthNodesMap.put(depth, list);
            }
            List<TreeNode> nodesList = depthNodesMap.get(depth);
            if (nodesList == null)
                return null;
            while (nodesList.size() > 1) {
                Set<TreeNode> set = new HashSet<TreeNode>();
                int size = nodesList.size();
                for (int i = 0; i < size; i++) {
                    TreeNode node = nodesList.remove(0);
                    TreeNode parent = childParentMap.get(node);
                    if (parent != null)
                        set.add(parent);
                }
                for (TreeNode node : set)
                    nodesList.add(node);
            }
            return nodesList.get(0);
        }
    }
    
  • ## Solution 1.
    
    
  • # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
        def subtreeWithAllDeepest(self, root):
            """
            :type root: TreeNode
            :rtype: TreeNode
            """
            return self.depth(root)[1]
            
        def depth(self, root):
            if not root: return 0, None
            l, r = self.depth(root.left), self.depth(root.right)
            if l[0] > r[0]:
                return l[0] + 1, l[1]
            elif l[0] < r[0]:
                return r[0] + 1, r[1]
            else:
                return l[0] + 1, root
            
    

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