1、二叉树的每个根节点到叶节点的路径代表一个数字，求所有根节点到叶节点所组成数字的和。
sum=12+13=25
1 int addsumnumber(BinTree *root,int sum) 2 { 3 if(root == NULL) 4 return 0; 5 if (root->left == 0 && root->right == 0) 6 return sum *10 + root->data; //左右子树都为空，则返回当前值 7 //递归遍历，将左右子树的值相加 8 return addsumnumber(root->left,sum *10 + root->data) + addsumnumber(root->right,sum *10 + root->data); 9 } 10 int sumnumbers(BinTree *root) 11 { 12 if(root == NULL) 13 return 0; 14 int sum = 0; 15 sum = addsumnumber(root,sum); 16 return sum; 17 }
2、计算二叉树的最大深度
1 int bintreemaxdepth(BinTree *root) 2 { 3 if(root == NULL) 4 return 0; 5 int ldepth = bintreemaxdepth(root->left); 6 int rdepth = bintreemaxdepth(root->right); 7 return max(ldepth,rdepth)+1; 8 }
3、计算二叉树的最小深度
1 int solution::MinDepth(TreeNode *root) 2 { 3 if (root ==NULL) 4 return 0; 5 int leftdepth = MinDepth(root->left); 6 int rightdepth = MinDepth(root->right); 7 if (leftdepth==0 && rightdepth==0) 8 return 1; 9 if (leftdepth==0) 10 leftdepth = INT_MAX; 11 if (rightdepth==0) 12 rightdepth = INT_MAX; 13 return min(leftdepth,rightdepth)+1; 14 }
4、二叉树中叶子节点个数
1 int bintreeleafnum(BinTree *root) 2 { 3 if(root == NULL) 4 return 0; 5 if(root->left == NULL && root->right == NULL) 6 return 1; 7 //总的叶子树=左子树的叶子树+右子树的叶子树 8 return bintreeleafnum(root->left) + bintreeleafnum(root->right); 9 }