104. Maximum Depth of Binary Tree
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}
}
class Solution {
int maxDepth;
public int maxDepth(TreeNode root) {
maxDepth = 0;
dfsHelper(root, 1);
return maxDepth;
}
private void dfsHelper(TreeNode root, int depth) {
if (root == null) {
return;
}
if (depth > maxDepth) {
maxDepth = depth;
}
dfsHelper(root.left, depth + 1);
dfsHelper(root.right, depth + 1);
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}
Deque<TreeNode> stack = new LinkedList<>();
Deque<Integer> depth = new LinkedList<>();
stack.push(root);
depth.push(1);
int maxDepth = 0;
while (!stack.isEmpty()) {
TreeNode curNode = stack.pollFirst();
int curDepth = depth.pollFirst();
maxDepth = Math.max(curDepth, maxDepth);
if (curNode.left != null) {
stack.offerFirst(curNode.left);
depth.offerFirst(curDepth + 1);
}
if (curNode.right != null) {
stack.offerFirst(curNode.right);
depth.offerFirst(curDepth + 1);
}
}
return maxDepth;
}
}
本解法本质就是 BFS 树的遍历,参考 102. Binary Tree Level Order Traversal
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}
Queue<TreeNode> queue = new LinkedList<>();
int maxDepth = 0;
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
if (cur.left != null) {
queue.offer(cur.left);
}
if (cur.right != null) {
queue.offer(cur.right);
}
}
maxDepth++;
}
return maxDepth;
}
}
所有解法 Time Complexity 均为 O(n),Space Complexity 为 O(h)。