LeetCode

235. 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).”

        _______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.

打算用后序遍历写的时候发现，题目是二叉搜索树。

所以根据二叉搜索树的特性，当pq两结点不在同一侧时，当前结点就一定是答案。时间复杂度O(logn)

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        if (!root) return NULL;        while (true) {            if (p->val < root->val && q->val < root->val)                root = root->left;            else if (p->val > root->val && q->val > root->val)                root = root->right;            else return root;        }    }};

236. Lowest Common Ancestor of a Binary Tree

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.

这次是在一颗二叉树中查寻最近公共祖先。

不得不说，二叉树搜索的时候用递归真的很优雅。只是每每想起有回面试官跟我说中根遍历写个非递归吧，我就心里一阵寒，就因为这样一直都比较排斥写递归，想用非递归写。然而这次还是写了递归，时间复杂度O(n)

其实用非递归写的时候，是最考验脑子糊不糊的时候。经常用递归很好写，一用非递归就各种出错。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        if (!root || root == q || root == p) return root;        TreeNode* le = lowestCommonAncestor(root->left, p, q);        TreeNode* ri = lowestCommonAncestor(root->right, p, q);        if (le && ri) return root;        return le ? le : ri;    }};