235. Lowest Common Ancestor of a Binary Search Tree

题目

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.

分析

所给的树是二分检索树，中根遍历得到由小到大的顺序排列，因此如果p,q 比root小, 则LCA必定在左子树, 如果p,q比root大, 则LCA必定在右子树. 如果一大一小,或者其中一个与root相等， 则root即为LCA。

/** * 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) {        TreeNode* temp;        if(root==NULL)            return NULL;        if(max(p->val, q->val) < root->val) {              return lowestCommonAncestor(root->left, p, q);          } else if(min(p->val, q->val) > root->val) {              return lowestCommonAncestor(root->right, p, q);          } else return root;     }};

假设给出的不是二分检索树，而是一个普通的二叉树，则需要分别从根结点出发，搜索p,q，保存两条路径，随后比较着两条路径，最后一个相同的结点即为LCA

/** * 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==null || p==null || q==null) return null;                List<TreeNode> pathp = new ArrayList<>();        List<TreeNode> pathq = new ArrayList<>();        pathp.add(root);        pathq.add(root);                getPath(root, p, pathp);        getPath(root, q, pathq);                TreeNode* lca = null;        for(int i=0; i<pathp.size() && i<pathq.size(); i++) {            if(pathp.get(i) == pathq.get(i)) lca = pathp.get(i);            else break;        }        return lca;    }        private boolean getPath(TreeNode* root, TreeNode* n, List<TreeNode> path) {        if(root==n) {            return true;        }                if(root.left!=null) {            path.add(root.left);            if(getPath(root.left, n, path)) return true;            path.remove(path.size()-1);        }                if(root.right!=null) {            path.add(root.right);            if(getPath(root.right, n, path)) return true;            path.remove(path.size()-1);        }                return false;    }}