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1123. Lowest Common Ancestor of Deepest Leaves

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:

img

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.

/**
 * 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) {
       if (root == null){
           return null;
       } 

       int left = height(root.left);
       int right = height(root.right);
       if (left == right){
           return root;
       }else if (left > right){
           return lcaDeepestLeaves(root.left);
       }else {
           return lcaDeepestLeaves(root.right);
       }
    }

    private int height(TreeNode root){
        if (root == null){
            return 0;
        }
        return 1 + Math.max(height(root.left), height(root.right));
    }
}

// TC: O(n)
// SC: O(n)

Test case explanation:

  1. [1,2,3]: 2 and 3 both leaves, so return 1, output [1,2,3] is the serialization of root 1
  2. [1,2,3,4]: 4 is the only leaf, so return 4 itself, output [4] is the serialization of leaf 4
  3. [1,2,3,4,5]: 4 and 5 both deepest leaves, so return their lca 2, output [2,4,5] is the serialization of node 2

so solution here is using dfs by always choosing the child with larger height until we find a node of both children with same height