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1123. Lowest Common Ancestor of Deepest Leaves

Description

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.

 

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.

 

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

Solutions

Solution 1: DFS

We design a function dfs(root) that returns a tuple (l, d), where l is the deepest common ancestor of node root, and d is the depth of node root. The execution logic of the function dfs(root) is as follows:

  • If root is null, return the tuple (None, 0);
  • Otherwise, we recursively call dfs(root.left) and dfs(root.right), obtaining tuples (l, d1) and (r, d2). If d1 > d2, the deepest common ancestor of root is l, and the depth is d1 + 1; if d1 < d2, the deepest common ancestor of root is r, and the depth is d2 + 1; if d1 = d2, the deepest common ancestor of root is root, and the depth is d1 + 1.

In the main function, we call dfs(root) and return the first element of its return value to get the deepest common ancestor node.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

  • /**
     * 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 lcaDeepestLeaves(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);
        }
    }
    
  • /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
     *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
     * };
     */
    class Solution {
    public:
        TreeNode* lcaDeepestLeaves(TreeNode* root) {
            return dfs(root).first;
        }
    
        pair<TreeNode*, int> dfs(TreeNode* root) {
            if (!root) {
                return {nullptr, 0};
            }
            auto [l, d1] = dfs(root->left);
            auto [r, d2] = dfs(root->right);
            if (d1 > d2) {
                return {l, d1 + 1};
            }
            if (d1 < d2) {
                return {r, d2 + 1};
            }
            return {root, d1 + 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 lcaDeepestLeaves(self, root: Optional[TreeNode]) -> Optional[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.
     * type TreeNode struct {
     *     Val int
     *     Left *TreeNode
     *     Right *TreeNode
     * }
     */
    type pair struct {
    	first  *TreeNode
    	second int
    }
    
    func lcaDeepestLeaves(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
    }
    
  • /**
     * Definition for a binary tree node.
     * class TreeNode {
     *     val: number
     *     left: TreeNode | null
     *     right: TreeNode | null
     *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
     *         this.val = (val===undefined ? 0 : val)
     *         this.left = (left===undefined ? null : left)
     *         this.right = (right===undefined ? null : right)
     *     }
     * }
     */
    
    function lcaDeepestLeaves(root: TreeNode | null): TreeNode | null {
        const dfs = (root: TreeNode | null): [TreeNode | null, number] => {
            if (root === null) {
                return [null, 0];
            }
            const [l, d1] = dfs(root.left);
            const [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];
    }
    
    

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