Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______3______
/
___5__ ___1__
/ /
6 _2 0 8
/
7 4
For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
题目含义：求一个普通二叉树中两个节点的最小公共父节点
1 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { 2 if (root == null || p == root || q == root) return root; 3 TreeNode left = lowestCommonAncestor(root.left, p, q); 4 TreeNode right = lowestCommonAncestor(root.right, p , q); 5 if (left!=null && right!=null) return root; //p和q与root的左右子节点相等，那么root即为p和q的公共父节点 6 return left!=null ? left : right;//p和q至少有一个为空，表示p和q落在了root的同一侧子树中，所以应该返回不为空的那个子树节点 7 }