回答編集履歴
1
補足を受けての追記
test
CHANGED
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@@ -1,3 +1,109 @@
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全ノードを、根から黒ノードの数を数えながら巡って、葉にたどり着いたら、黒接点の数を記録しておき、すでに記録した物と異なれば`0`を返して終了し、全部のノードを巡り終わったら`1`を返す。
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2
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3
3
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再帰呼び出しで巡れば難しくない気がします。
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#追記
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黒の数には、根と葉も入れるとして、こんな感じでしょうか。
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10
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11
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```C
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##include <stdio.h>
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typedef struct node { struct node *left, *right; int black; } NODE;
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int count_save;
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int check(NODE *this, int count);
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int main(void){
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NODE *root;
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NODE x1, x2, x3, x4, x5, x6;
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root = &x1;
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x1 = (NODE){ &x2, &x3, 0 };
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x2 = (NODE){ &x4, NULL, 1 };
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x3 = (NODE){ &x5, &x6, 0 };
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x4 = (NODE){ NULL, NULL, 0 };
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x5 = (NODE){ NULL, NULL, 1 };
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x6 = (NODE){ NULL, NULL, 1 };
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count_save = -1;
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printf("%d\n",check(root, 0));
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}
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int check(NODE *this, int count){
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count += this->black;
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if(!this->left && !this->right){ /* Leaf */
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if(count_save==-1){
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count_save = count;
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return 1;
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}else if(count_save != count){
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return 0;
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}else{
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return 1;
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}
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}
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if(this->left){
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if(check(this->left,count)==0){
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return 0;
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}
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}
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if(this->right){
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if(check(this->right,count)==0){
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return 0;
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}
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102
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103
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}
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104
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105
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return 1;
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106
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107
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}
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108
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109
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```
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