leetcode 111 求二叉树的最小深度
思路
采用广度优先遍历
代码
/**
* 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) {
if(root == null){
return 0;
}
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root); //将起点加入队列中
int step = 1;// 根节点本身就一层,所以初始化为1
while(!q.isEmpty()){
int size = q.size();
for(int i = 0; i<size; i++){
TreeNode currentNode = q.poll();
//判断是否达到终点
if(currentNode.left == null && currentNode.right == null){
return step;
}
if(currentNode.left != null){
q.offer(currentNode.left);
}
if(currentNode.right != null){
q.offer(currentNode.right);
}
}
step++;
}
return step;
}
}