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);
    }
}

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