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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.

  • /**
     * 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);
        }
    }
    
    ############
    
    /**
     * 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);
        }
    }
    
  • ## 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.
    # 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
            
    
  • /**
     * 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
    }
    

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