Formatted question description: https://leetcode.ca/all/1123.html

1123. Lowest Common Ancestor of Deepest Leaves (Medium)

Given the root of a binary tree, return the lowest common ancestor of its deepest leaves.

Recall that:

  • The node of a binary tree is a leaf if and only if it has no children
  • The depth of the root of the tree is 0. if the depth of a node is d, the depth of each of its children is d + 1.
  • The lowest common ancestor of a set S of nodes, is the node A with the largest depth such that every node in S is in the subtree with root A.

Note: This question is the same as 865: https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/

 

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 leaf-nodes of the tree.
Note that nodes 6, 0, and 8 are also leaf nodes, but the depth of them is 2, but the depth of nodes 7 and 4 is 3.

Example 2:

Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree, and it's the lca of itself.

Example 3:

Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest leaf node in the tree is 2, the lca of one node is itself.

 

Constraints:

  • The number of nodes in the tree will be in the range [1, 1000].
  • 0 <= Node.val <= 1000
  • The values of the nodes in the tree are unique.

Related Topics:
Tree, Depth-first Search

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/
// 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/lowest-common-ancestor-of-deepest-leaves/
// 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 lcaDeepestLeaves(TreeNode root) {
            List<TreeNode> deepestLeaves = new ArrayList<TreeNode>();
            Map<TreeNode, TreeNode> childParentMap = new HashMap<TreeNode, TreeNode>();
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.offer(root);
            while (!queue.isEmpty()) {
                deepestLeaves.clear();
                int size = queue.size();
                for (int i = 0; i < size; i++) {
                    TreeNode node = queue.poll();
                    TreeNode left = node.left, right = node.right;
                    if (left == null && right == null)
                        deepestLeaves.add(node);
                    if (left != null) {
                        childParentMap.put(left, node);
                        queue.offer(left);
                    }
                    if (right != null) {
                        childParentMap.put(right, node);
                        queue.offer(right);
                    }
                }
            }
            if (deepestLeaves.size() == 1)
                return deepestLeaves.get(0);
            Set<TreeNode> set = new HashSet<TreeNode>();
            for (TreeNode node : deepestLeaves) {
                TreeNode parent = childParentMap.get(node);
                set.add(parent);
            }
            while (set.size() > 1) {
                List<TreeNode> list = new ArrayList<TreeNode>(set);
                set.clear();
                for (TreeNode node : list) {
                    TreeNode parent = childParentMap.get(node);
                    set.add(parent);
                }
            }
            List<TreeNode> resultList = new ArrayList<TreeNode>(set);
            return resultList.get(0);
        }
    }
    
  • // OJ: https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/
    // 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);
        }
    };
    
  • # 1123. Lowest Common Ancestor of Deepest Leaves
    # https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/
    
    # 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 lcaDeepestLeaves(self, root: TreeNode) -> TreeNode:
            
            def helper(node):
                if not node: return 0, None
                
                h1, lca1 = helper(node.left)
                h2, lca2 = helper(node.right)
                
                if h1 > h2: return h1 + 1, lca1
                if h1 < h2: return h2 + 1, lca2
                
                return h1 + 1, node
            
            return helper(root)[1]
            
    
    

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