题目描述
思路
二叉树的最小深度就是第一个叶子节点所在的层数
方法一：前序遍历(递归、dfs)
/**
* 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 res = Integer.MAX_VALUE;
public int minDepth(TreeNode root) {
if(root == null) return 0;
dfs(root, 1);
return res;
}
private void dfs(TreeNode node, int level) {
if (node == null) return;
if (node.left == null && node.right == null) {
res = Math.min(res, level);
// 这里return可以写也可以不写,写的话相当于剪枝效果。
return;
}
dfs(node.left, level + 1);
dfs(node.right, level + 1);
}
}
方法二：层序遍历(bfs)
/**
* 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 minDepth(TreeNode root) {
Deque<TreeNode> queue = new ArrayDeque<>();
if (root == null) return 0;
queue.offer(root);
int level = 1;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i ++) {
TreeNode node = queue.poll();
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
// 二叉树的最小深度就是第一个叶子节点所在的层数
if (node.left == null && node.right == null) {
return level;
}
}
level += 1;
}
// 代码不会走到这里
throw new RuntimeException("Error!!!");
}
}