Microsoft interview questions: find the nearest common parent node of any two nodes on the binary tree.

 

Find the nearest common parent node of any two nodes on the binary tree.

 

 

 

It is required that the nearest common parent node (LCA, lowest common ancestor) of the two nodes can be traversed in descending order of the tree. If the two nodes are not on the same line, they must be on the left and right trees of node A respectively, node A is required to traverse to the first node that meets this condition in the descending order. In addition, when the two nodes are online, the requested node A is the parent node of the two nodes with the lowest levels.

Static bool lca (Node * root, int va, int vb, Node * & result, Node * parrent) <br/>{< br/> // left/right: whether the left/right subtree contains one of the two vertices to be judged <br/> bool left = false, right = false; <br/> if (! Result & root-> left) left = lca (root-> left, va, vb, result, root); <br/> if (! Result & root-> right) right = lca (root-> right, va, vb, result, root ); <br/> // whether the current node is one of the two vertices to be judged <br/> bool mid = false; <br/> if (root-> data = va | root-> data = vb) mid = true; <br/> if (! Result & int (left + right + mid) = 2) {<br/> if (mid) result = parrent; <br/> else result = root; <br/>}< br/> return left | mid | right; <br/>}</p> <p> Node * lca (Node * root, int va, int vb) <br/>{< br/> if (root = NULL) return NULL; <br/> Node * result = NULL; <br/> lca (root, va, vb, result, NULL); <br/> return result; <br/>}< br/>