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[LeetCode 236] Lowest Common Ancestor of a Binary Tree

Problem

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints:

  • The number of nodes in the tree is in the range [2, 10^5].
  • -10^9 <= Node.val <= 10^9
  • All Node.val are unique.
  • p != q
  • p and q will exist in the tree.

Thoughts

  • The node where it splits is the LCA
  • DFS left, right

TypeScript

/**
 * 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 lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: TreeNode | null): TreeNode | null {
    const dfs = (node: TreeNode): TreeNode => {
        if(node === null) {
            return null;
        }
        // descendant of itself
        if(node === p || node === q){
            return node;
        }
        
        const left: TreeNode = dfs(node.left);
        const right: TreeNode = dfs(node.right);
        
        // split
        if(left && right){
            return node;
        }
        // can't find
        if(!left && !right){
            return null
        }
        
        return left ? left : right;
    }
    
    return dfs(root)
};

Reference