1 //1 二叉树的深度(在后序遍历的基础上进行 时间复杂度: O(n)) (即左右子树的最大深度加1)
2 public int getDepth(BiTNode node)
3 {
4 if(node == null)
5 return 0;
6
7 int count1, count2;
8 count1 = getDepth(node.lchild);
9 count2 = getDepth(node.rchild);
10 return max(count1, count2) + 1;
11 }
1 //2 给定结点的后继(Successor)
2 public BiNode Successor(BiNode x)
3 {
4 BiNode p = x.rchild;
5 if(p != null)
6 {
7 while(p.lchild != null)
8 p = p.lchild;
9 return p;
10 }
11 BiNode q = x.parent;
12 while(q != null && x == q.rchild)
13 {
14 x = q;
15 q = q.parent;
16 }
17 return q;
18 }
1 //3 给定结点的前驱(Predecessor)
2 public BiNode Predecessor(BiNode x)
3 {
4 BiNode p = x.lchild;
5 if(p != null)
6 {
7 while(p.rchild != null)
8 p = p.rchild;
9 return p;
10 }
11 BiNode q = x.parent;
12 while(q != null && x == q.lchild)
13 {
14 x = q;
15 q = q.parent;
16 }
17 return q;
18 }