最低公共祖先 lowest common ancestor

最低公共祖先 lowest common ancestor

The idea is to traverse the tree starting from root. If any of the given keys (n1 and n2) matches with root, then root is LCA (assuming that both
keys are present). If root doesn’t match with any of the keys, we recur for left and right subtree. The node which has one key present in its left subtree and the other key present in right subtree is the LCA. If both keys lie in left subtree, then left subtree
has LCA also, otherwise LCA lies in right subtree.

下面给出的算法的假设是，两个输入值都存在于二叉树中。如果有一个在，一个不在，那么函数返回值是那个存在的节点。

一个完整无误的算法，请看：http://www.geeksforgeeks.org/lowest-common-ancestor-binary-tree-set-1/

/* Program to find LCA of n1 and n2 using one traversal of Binary Tree */
#include <iostream>
using namespace std;

// A Binary Tree Node
struct Node
{
struct Node *left, *right;
int key;
};

// Utility function to create a new tree Node
Node* newNode(int key)
{
Node *temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return temp;
}

// This function returns pointer to LCA of two given values n1 and n2.
// This function assumes that n1 and n2 are present in Binary Tree
struct Node *findLCA(struct Node* root, int n1, int n2)
{
// Base case
if (root == NULL) return NULL;

// If either n1 or n2 matches with root's key, report
// the presence by returning root (Note that if a key is
// ancestor of other, then the ancestor key becomes LCA
if (root->key == n1 || root->key == n2)
return root;

// Look for keys in left and right subtrees
Node *left_lca  = findLCA(root->left, n1, n2);
Node *right_lca = findLCA(root->right, n1, n2);

// If both of the above calls return Non-NULL, then one key
// is present in once subtree and other is present in other,
// So this node is the LCA
if (left_lca && right_lca)  return root;

// Otherwise check if left subtree or right subtree is LCA
return (left_lca != NULL)? left_lca: right_lca;
}

// Driver program to test above functions
int main()
{
// Let us create binary tree given in the above example
Node * root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
cout << "LCA(4, 5) = " << findLCA(root, 4, 5)->key;
cout << "\nLCA(4, 6) = " << findLCA(root, 4, 6)->key;
cout << "\nLCA(3, 4) = " << findLCA(root, 3, 4)->key;
cout << "\nLCA(2, 4) = " << findLCA(root, 2, 4)->key;
return 0;
}