# Lowest common ancestor of a binary tree Part I

Source: Internet
Author: User

`456789101112131415161718192021// Return #nodes that matches P or Q in the subtree.int countMatchesPQ(Node *root, Node *p, Node *q) {  if (!root) return 0;  int matches = countMatchesPQ(root->left, p, q) + countMatchesPQ(root->right, p, q);  if (root == p || root == q)    return 1 + matches;  else    return matches;} Node *LCA(Node *root, Node *p, Node *q) {  if (!root || !p || !q) return NULL;  if (root == p || root == q) return root;  int totalMatches = countMatchesPQ(root->left, p, q);  if (totalMatches == 1)    return root;  else if (totalMatches == 2)    return LCA(root->left, p, q);  else /* totalMatches == 0 */    return LCA(root->right, p, q);}`

Bottom to up. Post-order traversal

`Node *LCA(Node *root, Node *p, Node *q) {  if (!root) return NULL;  if (root == p || root == q) return root;  Node *L = LCA(root->left, p, q);  Node *R = LCA(root->right, p, q);  if (L && R) return root;  // if p and q are on both sides  return L ? L : R;  // either one of p,q is on one side OR p,q is not in L&R subtrees}`

Lowest common ancestor of a binary tree Part I

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