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236. 二叉树的最近公共祖先 #93
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原题链接
递归 dfs
1.从根节点开始遍历,递归左右子树
2.递归终止条件:当前节点为空或者等于 p 或 q,返回当前节点
3.p,q 可能在相同的子树中,也可能在不同的子树中
4.如果左右子树查到节点都不为空,则表示 p 和 q 分别在左右子树中,当前节点就是最近公共祖先
5.如果左右子树中有一个不为空,则返回为空节点
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