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
        

865 - Smallest Subtree with all the Deepest Nodes

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

Solution 1.

Solution 2.

All Problems

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