Lowest Common Ancestor of a Binary Search Tree

最新更新请见：https://yanjia.me/zh/2019/02/…

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

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

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

二分法

复杂度

时间 O(h) 空间 O(h) 递归栈空间

思路

对于二叉搜索树，公共祖先的值一定大于等于较小的节点，小于等于较大的节点。换言之，在遍历树的时候，如果当前结点大于两个节点，则结果在当前结点的左子树里，如果当前结点小于两个节点，则结果在当前节点的右子树里。

代码

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q);
        if(root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q);
        return root;
    }
}

Lowest Common Ancestor of a Binary Tree

最新更新请见：https://yanjia.me/zh/2019/02/…

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

深度优先标记

复杂度

时间 O(h) 空间 O(h) 递归栈空间

思路

我们可以用深度优先搜索，从叶子节点向上，标记子树中出现目标节点的情况。如果子树中有目标节点，标记为那个目标节点，如果没有，标记为null。显然，如果左子树、右子树都有标记，说明就已经找到最小公共祖先了。如果在根节点为p的左右子树中找p、q的公共祖先，则必定是p本身。

换个角度，可以这么想：如果一个节点左子树有两个目标节点中的一个，右子树没有，那这个节点肯定不是最小公共祖先。如果一个节点右子树有两个目标节点中的一个，左子树没有，那这个节点肯定也不是最小公共祖先。只有一个节点正好左子树有，右子树也有的时候，才是最小公共祖先。

代码

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        //发现目标节点则通过返回值标记该子树发现了某个目标结点
        if(root == null || root == p || root == q) return root;
        //查看左子树中是否有目标结点，没有为null
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        //查看右子树是否有目标节点，没有为null
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        //都不为空，说明做右子树都有目标结点，则公共祖先就是本身
        if(left!=null&&right!=null) return root;
        //如果发现了目标节点，则继续向上标记为该目标节点
        return left == null ? right : left;
    }
}