Starting from root node and moving downwards if you find any node that has either p or q as its direct child then it is the LCA. (edit – this should be if p or q is the node’s value, return it. Otherwise it will fail when one of p or q is a direct child of the other.)
Else if you find a node with p in its right(or left) subtree and q in its left(or right) subtree then it is the LCA.
The fixed code looks like:
treeNodePtr findLCA(treeNodePtr root, treeNodePtr p, treeNodePtr q) {
// no root no LCA.
if(!root) {
return NULL;
}
// if either p or q is the root then root is LCA.
if(root==p || root==q) {
return root;
} else {
// get LCA of p and q in left subtree.
treeNodePtr l=findLCA(root->left , p , q);
// get LCA of p and q in right subtree.
treeNodePtr r=findLCA(root->right , p, q);
// if one of p or q is in leftsubtree and other is in right
// then root it the LCA.
if(l && r) {
return root;
}
// else if l is not null, l is LCA.
else if(l) {
return l;
} else {
return r;
}
}
}
The below code fails when either is the direct child of other.
treeNodePtr findLCA(treeNodePtr root, treeNodePtr p, treeNodePtr q) {
// no root no LCA.
if(!root) {
return NULL;
}
// if either p or q is direct child of root then root is LCA.
if(root->left==p || root->left==q ||
root->right ==p || root->right ==q) {
return root;
} else {
// get LCA of p and q in left subtree.
treeNodePtr l=findLCA(root->left , p , q);
// get LCA of p and q in right subtree.
treeNodePtr r=findLCA(root->right , p, q);
// if one of p or q is in leftsubtree and other is in right
// then root it the LCA.
if(l && r) {
return root;
}
// else if l is not null, l is LCA.
else if(l) {
return l;
} else {
return r;
}
}
}