学步园

TreeNode findLCA(TreeNode root, TreeNode p, TreeNode 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.
                TreeNode l=findLCA(root->left , p , q);

                // get LCA of p and q in right subtree.
                TreeNode 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;
                }
        }
}

这个面试太常出现了，所以更新下。最主要用buttom up效率高，O(n), 思路其实类似isBalanced Binary Tree.