Leetcode-235. Lowest Common Ancestor of a Binary Search Tree

一种复杂的方法：首先将从根节点到p,q两节点的路径上的结点压入栈中，处理两个栈使栈容量保持一致，同时弹出两栈中的元素，直至两栈栈顶元素相同。

class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        stack<TreeNode*> s1,s2;        s1.push(root);        do{            if(s1.top()->val < p->val)                s1.push(s1.top()->right);            else if (s1.top()->val > p->val)                s1.push(s1.top()->left);        }while(s1.top()!=p);                s2.push(root);        do{            if(s2.top()->val < q->val)                s2.push(s2.top()->right);            else if (s2.top()->val > q->val)                s2.push(s2.top()->left);        } while(s2.top()!=q);                        while(s1.size()>s2.size())            s1.pop();        while(s2.size()>s1.size())            s2.pop();                    while(s1.top() != s2.top()){            s1.pop();            s2.pop();        }            return s1.top();    }};

一种非常简单的思路

Just walk down from the whole tree's root as long as both p and q are in the same subtree (meaning their values are both smaller or both larger than root's). This walks straight from the root to the LCA, not looking at the rest of the tree, so it's pretty much as fast as it gets. A few ways to do it:

Iterative, O(1) space

Python

def lowestCommonAncestor(self, root, p, q):    while (root.val - p.val) * (root.val - q.val) > 0:        root = (root.left, root.right)[p.val > root.val]    return root

Java

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {    while ((root.val - p.val) * (root.val - q.val) > 0)        root = p.val < root.val ? root.left : root.right;    return root;}

(in case of overflow, I'd do (root.val - (long)p.val) * (root.val - (long)q.val))

Different Python

def lowestCommonAncestor(self, root, p, q):    a, b = sorted([p.val, q.val])    while not a <= root.val <= b:        root = (root.left, root.right)[a > root.val]    return root

"Long" Python, maybe easiest to understand

def lowestCommonAncestor(self, root, p, q):    while root:        if p.val < root.val > q.val:            root = root.left        elif p.val > root.val < q.val:            root = root.right        else:            return root

Recursive

Python

def lowestCommonAncestor(self, root, p, q):    next = p.val < root.val > q.val and root.left or \           p.val > root.val < q.val and root.right    return self.lowestCommonAncestor(next, p, q) if next else root

Python One-Liner

def lowestCommonAncestor(self, root, p, q):    return root if (root.val - p.val) * (root.val - q.val) < 1 else \           self.lowestCommonAncestor((root.left, root.right)[p.val > root.val], p, q)

Java One-Liner

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {    return (root.val - p.val) * (root.val - q.val) < 1 ? root :           lowestCommonAncestor(p.val < root.val ? root.left : root.right, p, q);}

"Long" Python, maybe easiest to understand

def lowestCommonAncestor(self, root, p, q):    if p.val < root.val > q.val:        return self.lowestCommonAncestor(root.left, p, q)    if p.val > root.val < q.val:        return self.lowestCommonAncestor(root.right, p, q)    return root