Lowest Common Ancestor of a Binary Tree
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).”
Divide and Conquer
Time Complexity
O(N)思路
首先要先确定给的两个node是否都在tree里，如果都在tree里的话，就可以分成3种情况，第一种情况是两个节点是在公共祖先的左右两侧，第二种情况是都在树的左侧，第三种情况是都在树的右侧，如果是第二，第三种情况的话，公共祖先就在给定的两个点中比较上面的那一个。
如果转换成代码的话，从上往下走，base case分为3种，判断遇到了p就直接返回p，遇到q就直接返回q，不用向下做了。如果left,right都不为空，就返回root自己；left,right哪一个不为空就返回哪个，整个recursion做完就可以得到LCA。
代码
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root == null) return null; if(root == p) return p; if(root == q) return q; TreeNode left = lowestCommonAncestor(root.left, p, q); TreeNode right = lowestCommonAncestor(root.right, p, q); if(left != null && right != null) return root; return left != null ? left : right;} Lowest Common Ancestor of a Binary Search Tree
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).”
Divide and Conquer
Time Complexity
O(N)思路
LCA肯定是落在p,q两个点之间的，或者是p, q其中一个点，在BST中，这就等于LCA的大小一定是在p.val,q.val之间的，或者等于p.val, q.val。从上往下走，遇到的第一个p,q之间的或者是p,q其中一个，一定是LCA，不用再继续往下走了。如果cur root，比p,q中的都大的话，cur往左走，比p,q中的都小的话cur往右走。
代码
Recursive
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root == null) return null; 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;} Non-Recursive
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root == null) return null; while(true){ if(root.val > p.val && root.val > q.val){ root = root.left; }else if(root.val < p.val && root.val < q.val){ root = root.right; }else break; } return root;} Lowest Common Ancestor of a Binary Tree With Parent Node
思路
如果有Parent指针的话，这就又变成Linkedlist求intersection了