103 lines
3.4 KiB
C++
103 lines
3.4 KiB
C++
// Source : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
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// Author : Hao Chen
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// Date : 2015-07-17
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/**********************************************************************************
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*
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* Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the
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* tree.
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*
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* According to the definition of LCA on Wikipedia: “The lowest common ancestor is
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* defined between two nodes v and w as the lowest node in T that has both v and w as
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* descendants (where we allow a node to be a descendant of itself).”
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*
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* _______3______
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* / \
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* ___5__ ___1__
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* / \ / \
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* 6 _2 0 8
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* / \
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* 7 4
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*
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* For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example
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* is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according
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* to the LCA definition.
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*
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*
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*
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**********************************************************************************/
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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bool findPath(TreeNode* root, TreeNode* p, vector<TreeNode*>& path) {
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if (root==NULL) return false;
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if (root == p) {
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path.push_back(p);
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return true;
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}
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path.push_back(root);
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if (findPath(root->left, p, path)) return true;
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if (findPath(root->right, p, path)) return true;
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path.pop_back();
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return false;
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}
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//Ordinary way, find the path and comapre the path.
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TreeNode* lowestCommonAncestor01(TreeNode* root, TreeNode* p, TreeNode* q) {
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vector<TreeNode*> path1, path2;
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if (!findPath(root, p, path1)) return NULL;
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if (!findPath(root, q, path2)) return NULL;
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int len = path1.size() < path2.size() ? path1.size() : path2.size();
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TreeNode* result = root;
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for(int i=0; i<len; i++) {
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if (path1[i] != path2[i]) {
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return result;
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}
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result = path1[i];
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}
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return result;
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}
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//Actually, we can do the recursive search in one time
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TreeNode* lowestCommonAncestor02(TreeNode* root, TreeNode* p, TreeNode* q) {
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//return if found or not found, return NULL if not found
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if (root==NULL || root == p || root == q) return root;
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//find the LCA in left tree
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TreeNode* left = lowestCommonAncestor02(root->left, p, q);
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//find the LCA in right tree
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TreeNode* right = lowestCommonAncestor02(root->right, p, q);
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//left==NULL means both `p` and `q` are not found in left tree.
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if (left==NULL) return right;
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//right==NULL means both `p` and `q` are not found in right tree.
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if (right==NULL) return left;
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// left!=NULL && right !=NULL, which means `p` & `q` are seperated in left and right tree.
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return root;
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}
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TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
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srand(time(0));
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if (random()%2) {
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return lowestCommonAncestor02(root, p, q);
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}
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return lowestCommonAncestor01(root, p, q);
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}
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};
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