题解 | #在二叉树中找到两个节点的最近公共祖先#

/*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)
 *     }
 * }
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param root TreeNode类 
 * @param o1 int整型 
 * @param o2 int整型 
 * @return int整型
 */
export function lowestCommonAncestor(root: TreeNode, o1: number, o2: number): number {
    if(root == null){
        return null
    }
    if(root.val == o1  || root.val == o2){
        return root.val
    }
    const  left  = lowestCommonAncestor(root.left,o1,o2)
    const right = lowestCommonAncestor(root.right,o1,o2)
    if(left ==null&&right == null)return null
    if(left !== null && right !== null)return root.val
    return left ? left:right
}