Maximum Depth of Binary Tree
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Question Solution
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
我的解决方案：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxDepth(TreeNode* root)
{
if(NULL == root)
return 0;
int depth_l = maxDepth(root->left);
int depth_r = maxDepth(root->right);
return depth_l > depth_r ? depth_l + 1:depth_r + 1;
}
};一行代码的解法：
int maxDepth(TreeNode *root)
{
return root == NULL ? 0 : max(maxDepth(root -> left), maxDepth(root -> right)) + 1;
}
不用递归的解法：Breadth-first-search
int maxDepth(TreeNode *root)
{
if(root == NULL)
return 0;
int res = 0;
queue<TreeNode *> q;
q.push(root);
while(!q.empty())
{
++ res;
for(int i = 0, n = q.size(); i < n; ++ i)
{
TreeNode *p = q.front();
q.pop();
if(p -> left != NULL)
q.push(p -> left);
if(p -> right != NULL)
q.push(p -> right);
}
}
return res;
}
不用递归的解法2
int maxDepth(TreeNode *root)
{
if (root == NULL) return 0;
stack<TreeNode *> gray;
stack<int> depth;
int out = 0;
gray.push(root);
depth.push(1);
while (!gray.empty()) {
TreeNode *tmp = gray.top();
int num = depth.top();
gray.pop();
depth.pop();
if (tmp->left == NULL && tmp->right == NULL) {
out = num > out ? num : out;
}
else {
if (tmp->left != NULL) {
gray.push(tmp->left);
depth.push(num + 1);
}
if (tmp->right != NULL) {
gray.push(tmp->right);
depth.push(num + 1);
}
}
}
return out;
}
python 的解决方案：
# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# @param root, a tree node
# @return an integer
def maxDepth(self, root):
def maxDepthHelper(root):
if not root: return 0
return max(1+maxDepthHelper(root.left), 1+maxDepthHelper(root.right))
return maxDepthHelper(root)