Question :
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.
Anwser 1 : DFS
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxDepth(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root == NULL) return 0;
int l = maxDepth(root->left);
int r = maxDepth(root->right);
return l > r ? l + 1 : r + 1;
}
};
More simpe :
class Solution {
public:
int maxDepth(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root == NULL) return 0;
return 1 + max( maxDepth(root->left), maxDepth(root->right) );
}
};
Anwser 2 : BFS in queue
class Solution {
public:
int maxDepth(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root == NULL) return 0;
queue<TreeNode *> Q;
Q.push(root);
int count = 1;
int depth = 0;
while(!Q.empty()){
TreeNode *tmp = Q.front();
Q.pop();
count--;
if(tmp->left){
Q.push(tmp->left);
}
if(tmp->right){
Q.push(tmp->right);
}
if(count == 0){
depth++; // one row is end, add a depth
count = Q.size(); // next row count
}
}
return depth;
}
};
Anwser 3 : BFS in stack
class Solution {
public:
int maxDepth(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root == NULL) return 0;
stack<TreeNode*> S;
int maxDepth = 0;
TreeNode *prev = NULL;
S.push(root);
while (!S.empty()) {
TreeNode *curr = S.top();
if (prev == NULL || prev->left == curr || prev->right == curr) {
if (curr->left)
S.push(curr->left);
else if (curr->right)
S.push(curr->right);
} else if (curr->left == prev) {
if (curr->right)
S.push(curr->right);
} else {
S.pop();
}
prev = curr;
if (S.size() > maxDepth)
maxDepth = S.size();
}
return maxDepth;
}
};
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