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前言

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如果说笔试的时候经常遇到各种动归回溯这类稍有难度的题目，那么面试会倾向于一些比较经典的问题，难度不算大，而且也比较实用。 本文就用 Git 引出一个经典的算法问题：最近公共祖先（Lowest Common Ancestor，简称 LCA）。 git pull 这个命令我们经常会用，它默认是使用 merge 方式将远端别人的修改拉到本地；如果带上参数 git pull -r，就会使用 rebase 的方式将远端修改拉到本地。 这二者最直观的区别就是：merge 方式合并的分支会看到很多「分叉」，而 rebase 方式合并的分支就是一条直线。但无论哪种方式，如果存在冲突，Git 都会检测出来并让你手动解决冲突。 那么问题来了，Git 是如何检测两条分支是否存在冲突的呢？

一、力扣236. 二叉树的最近公共祖先

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {return fun(root,p.val,q.val);}public TreeNode fun(TreeNode root ,int val1, int val2){if(root == null){return null;}if(root.val == val1 || root.val == val2){return root;}TreeNode lchild = fun(root.left, val1, val2);TreeNode rchild = fun(root.right, val1, val2);if(lchild != null && rchild != null){return root;}return lchild != null ? lchild : rchild;}
}

二、力扣1676. 二叉树的最近公共祖先 IV

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode[] nodes) {if(root == null){return null;}for(TreeNode n : nodes){if(root == n){return root;}}TreeNode lchild = lowestCommonAncestor(root.left,nodes);TreeNode rchild = lowestCommonAncestor(root.right,nodes);if(lchild != null && rchild != null){return root;}return lchild != null ? lchild : rchild;}
}

三、力扣1644. 二叉树的最近公共祖先 II

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {boolean flagL = false, flagR = false;public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {TreeNode res = fun(root,p,q);if(flagL && flagR){return res;}return null;}public TreeNode fun(TreeNode root, TreeNode p, TreeNode q){if(root == null){return null;}TreeNode lchild = fun(root.left,p,q);TreeNode rchild = fun(root.right, p, q);if(lchild != null && rchild != null){return root;}if(root == p || root == q){if(root == p){flagL = true;return root;}if(root == q){flagR = true;return root;}}return lchild != null ? lchild : rchild; }
}

四、力扣235. 二叉搜索树的最近公共祖先

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/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);}else if(root.val < p.val &&root.val < q.val){return lowestCommonAncestor(root.right,p,q);}else{return root;}}
}

五、力扣1650. 二叉树的最近公共祖先 III

/*
// Definition for a Node.
class Node {public int val;public Node left;public Node right;public Node parent;
};
*/class Solution {public Node lowestCommonAncestor(Node p, Node q) {while(fun(p,q) == null){p = p.parent;}return p;}public Node fun(Node p, Node q){if(p == null){return null;}if(p == q){return p;}Node l = fun(p.left,q);Node r = fun(p.right,q);if(l != null){return l;}if(r != null){return r;}return null;}
}

链表寻找共同节点操作

/*
// Definition for a Node.
class Node {public int val;public Node left;public Node right;public Node parent;
};
*/class Solution {public Node lowestCommonAncestor(Node p, Node q) {Node a = p, b = q;while(a != b){if(a == null) a = q;else a = a.parent;if(b == null) b = p;else b = b.parent;}return a;}
}