22_104. 二叉树的最大深度
题目描述：
解题思路：
- 自底向上：当知道某个节点的左子树和右子树的最大深度时，就可以知道这棵树的最大深度是
max(leftDepth,rightDepth) + 1；- 自顶向下：可以把深度当做参数传递到下一层，当递归到叶子节点时，更新最大深度，否则就是
maxDepth(root.left,depth+1),maxDepth(root.right,depth+1)；- 广度优先搜索：利用队列，广度优先搜索，记录层数
ans，每次遍历到下一层时，更新ans，利用queue.size来控制将整个队列的节点先出队，再将其子节点入队。
代码：
自底向上
//自底向上：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private int max = 0;
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}
int leftDepth = maxDepth(root.left);
int rightDepth = maxDepth(root.right);
max = Math.max(leftDepth, rightDepth) + 1;
return max;
}
}
自顶向下
//自顶向下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private int answer = 0;
public int maxDepth(TreeNode root) {
maxDepth(root, 1);
return answer;
}
public void maxDepth(TreeNode root, int depth) {
if (root == null) {
return;
}
if (root.left == null && root.right == null) {
answer = Math.max(answer, depth);
}
maxDepth(root.left, depth + 1);
maxDepth(root.right, depth + 1);
}
}
广度优先搜索
//广度优先搜索：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}
int ans = 0;
Queue queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()) {
int size = queue.size();
while (size > 0) {
TreeNode node = queue.poll();
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
size--;
}
ans++;
}
return ans;
}
}