我认为最优雅的解决方案就是这个。是的,当然我有偏见。我是人类:-)
def countLeft (node,ind):
if node == null: return 0
return ind + countLeft (node->left, 1) + countLeft (node->right, 0)
total = countLeft (root, 0)
通过向下传递左节点的指标,它简化了必须传递的内容。下图显示了每个传递的总和 - 您从底部开始,每个null都递增0.
左侧的每个节点都会传递1以及来自两个分支的任何节点。右侧的每个节点都会传递0加上来自两个分支的任何节点。
根本没有增加任何东西,因为它既不是左边节点也不是右边节点(它的处理方式与右边相同)。
4
^
|
+---+
| 3 |
__________+---+__________
/2 2\
+---+ +---+
| 5 | | 2 |
+---+ +---+
/1 /2 0\
+---+ +---+ +---+
| 1 | | 4 | | 6 |
+---+ +---+ +---+
/0 0\ /1 0\ /0 0\
+---+
| 7 |
+---+
/0 0\
你可以从这个完整的程序见操作:
#include <stdio.h>
typedef struct sNode { int val; struct sNode *left, *right; } tNode;
#define setNode(N,V,L,R) N.val = V; N.left = L; N.right = R
int countLeft (tNode *node, int ind) {
if (node == NULL) return 0;
int x = ind + countLeft (node->left, 1) + countLeft (node->right, 0);
printf ("Node %d passing up %d\n", node->val, x);
return x;
}
int main (void) {
tNode n3, n5, n1, n2, n4, n6, n7;
setNode (n3, 3, &n5, &n2);
setNode (n5, 5, &n1, NULL);
setNode (n1, 1, NULL, NULL);
setNode (n2, 2, &n4, &n6);
setNode (n4, 4, &n7, NULL);
setNode (n7, 7, NULL, NULL);
setNode (n6, 6, NULL, NULL);
printf ("countLeft is %d\n", countLeft (&n3, 0));
return 0;
}
,它输出的调试线路:
Node 1 passing up 1
Node 5 passing up 2
Node 7 passing up 1
Node 4 passing up 2
Node 6 passing up 0
Node 2 passing up 2
Node 3 passing up 4
countLeft is 4
的countLeft
函数的非调试版本就像这个答案开始时的伪代码一样简单:
int countLeft (tNode *node, int ind) {
if (node == NULL) return 0;
return ind + countLeft (node->left, 1) + countLeft (node->right, 0);
}
它会不会命中空指针exxeption? – Catie 2010-11-02 09:05:59
@Catie:不,因为它所做的第一件事是检查null,然后在这种情况下返回0。这是执行此操作的标准方法,因此您不必将根节点视为特殊节点。 – paxdiablo 2010-11-02 09:13:11