题目描述
- 输入一棵二叉树的根结点,求该树的深度。从根结点到叶结点依次经过的结点(含根、叶结点)形成树的一条路径,最长路径的长度为树的深度。
算法分析
- 法一:前序遍历;
- 法二:如果该树只有一个结点,它的深度为1;如果根节点只有左子树没有右子树,那么树的深度为左子树的深度加1;同样,如果只有右子树没有左子树,那么树的深度为右子树的深度加1。如果既有左子树也有右子树,那该树的深度就是左子树和右子树的最大值加1.
提交代码:
class Solution {
public:
int TreeDepth(TreeNode* pRoot)
{
if (!pRoot)
return 0;
int left = TreeDepth(pRoot->left);
int right = TreeDepth(pRoot->right);
return left > right ? left + 1 : right + 1;
}
// 前序遍历
int TreeDepth2(TreeNode* pRoot)
{
if (!pRoot)
return 0;
int max = 0, curr = 0;
TreeDepthCore(pRoot, curr, &max);
return max;
}
void TreeDepthCore(TreeNode* pRoot, int curr, int *max)
{
if (!pRoot)
{
if (curr > *max)
*max = curr;
return;
}
// 前序遍历
++curr;
TreeDepthCore(pRoot->left, curr, max);
TreeDepthCore(pRoot->right, curr, max);
}
};
测试代码:
// ====================测试代码====================
void Test(const char* testName, TreeNode* pRoot, int expected)
{
Solution s;
int result = s.TreeDepth(pRoot);
if (expected == result)
printf("%s passed.\n", testName);
else
printf("%s FAILED.\n", testName);
}
// 1
// / \
// 2 3
// /\ \
// 4 5 6
// /
// 7
void Test1()
{
TreeNode* pNode1 = CreateBinaryTreeNode(1);
TreeNode* pNode2 = CreateBinaryTreeNode(2);
TreeNode* pNode3 = CreateBinaryTreeNode(3);
TreeNode* pNode4 = CreateBinaryTreeNode(4);
TreeNode* pNode5 = CreateBinaryTreeNode(5);
TreeNode* pNode6 = CreateBinaryTreeNode(6);
TreeNode* pNode7 = CreateBinaryTreeNode(7);
ConnectTreeNodes(pNode1, pNode2, pNode3);
ConnectTreeNodes(pNode2, pNode4, pNode5);
ConnectTreeNodes(pNode3, nullptr, pNode6);
ConnectTreeNodes(pNode5, pNode7, nullptr);
Test("Test1", pNode1, 4);
DestroyTree(pNode1);
}
// 1
// /
// 2
// /
// 3
// /
// 4
// /
// 5
void Test2()
{
TreeNode* pNode1 = CreateBinaryTreeNode(1);
TreeNode* pNode2 = CreateBinaryTreeNode(2);
TreeNode* pNode3 = CreateBinaryTreeNode(3);
TreeNode* pNode4 = CreateBinaryTreeNode(4);
TreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, pNode2, nullptr);
ConnectTreeNodes(pNode2, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode5, nullptr);
Test("Test2", pNode1, 5);
DestroyTree(pNode1);
}
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void Test3()
{
TreeNode* pNode1 = CreateBinaryTreeNode(1);
TreeNode* pNode2 = CreateBinaryTreeNode(2);
TreeNode* pNode3 = CreateBinaryTreeNode(3);
TreeNode* pNode4 = CreateBinaryTreeNode(4);
TreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, nullptr, pNode2);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode5);
Test("Test3", pNode1, 5);
DestroyTree(pNode1);
}
// 树中只有1个结点
void Test4()
{
TreeNode* pNode1 = CreateBinaryTreeNode(1);
Test("Test4", pNode1, 1);
DestroyTree(pNode1);
}
// 树中没有结点
void Test5()
{
Test("Test5", nullptr, 0);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
return 0;
}