题目描述
Given a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
采用层次遍历非递归方法(使用队列)
public class Solution {
public int run(TreeNode root) {
//采用非递归,层次遍历,找叶节点
int minDepth = 0;
//若根节点为null,则返回;
if(root == null)
return minDepth;
LinkedList<TreeNode> queue = new LinkedList<>();
//last记录没层的最后一个节点,第一层的最后一个节点即根节点
TreeNode last = root;
//将根节点加入queue,minDepth++
queue.add(root);
minDepth++;
//判断条件,是否为空
while(!queue.isEmpty()){
//将队首节点弹出并删除
root = queue.poll();
//如果该节点没有子节点则返回
if(root.left == null && root.right == null)
return minDepth;
//若左子树非空,或右子树非空,则将该节点的子节点入队
if(root.left != null){
queue.add(root.left);
}
if(root.right != null){
queue.add(root.right);
}
//若该节点是该层的最后一个节点,则minDepth++,并获取下一层的最后一个节点
if(root == last){
minDepth ++;
last = queue.getLast();
}
}
return minDepth;
}
}
采用深度优先,递归
public class Solution {
public int run(TreeNode root) {
if(root == null)
return 0;
if(root.left == null)
return 1 + run(root.right);
if(root.right == null)
return 1 + run(root.left);
return 1 + Math.min(run(root.right),run(root.left));
}
}