Leecode 111.二叉树的最小深度
/*
* 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) {
//先判断是否为空
if(root == null) {
return 0;
}
int depth=0;
Queue<TreeNode> que = new LinkedList<>();
que.offer(root);
while (!que.isEmpty()) {
int len = que.size();
depth ++;
for (int i = 0; i < len; i++) {
TreeNode node = que.poll();
//判断两个子节点都为空则为叶子节点
if(node.right == null && node.left == null ){
return depth;
}
if(node.left != null) {que.offer(node.left);}
if(node.right != null) {que.offer(node.right);}
}
}
return depth;
}
}
//递归法
/**
* 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) {
//先判断是否为空
if(root == null) {
return 0;
}
//递归法
int leftLenth = minDepth(root.left);
int rightLenth = minDepth(root.right);
//从处理中间节点
//如果左节点为空,右不为空,说明不是最小深度
if(root.left == null && root.right != null) {
return rightLenth+1;
}
if(root.right == null && root.left != null) {
return leftLenth +1;
}
return min(leftLenth,rightLenth) +1;
}
public int min(int leftLenth,int rightLenth) {
return leftLenth < rightLenth ?leftLenth:rightLenth;
}
}
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