111. Minimum Depth of Binary Tree
原题链接:https://leetcode.com/problems/minimum-depth-of-binary-tree/description/
简单级别的题目就是不咋滴难哦:
/**
* Created by clearbug on 2018/2/26.
*/
public class Solution {
static class TreeNode {
int val;
TreeNode left;
TreeNode right;
public TreeNode(int val) {
this.val = val;
}
}
public static void main(String[] args) {
Solution s = new Solution();
// test1
TreeNode root = new TreeNode(3);
root.left = new TreeNode(9);
root.right = new TreeNode(20);
root.right.left = new TreeNode(15);
root.right.right = new TreeNode(7);
System.out.println(s.minDepth(root));
// test2
TreeNode root2 = new TreeNode(1);
root2.left = new TreeNode(2);
root2.right = new TreeNode(2);
root2.left.left = new TreeNode(3);
root2.left.right = new TreeNode(3);
root2.left.left.left = new TreeNode(4);
root2.left.left.right = new TreeNode(4);
System.out.println(s.minDepth(root2));
// test3
TreeNode root3 = new TreeNode(1);
root3.left = new TreeNode(2);
root3.right = new TreeNode(2);
root3.left.left = new TreeNode(3);
root3.left.right = new TreeNode(3);
root3.right.left = new TreeNode(3);
root3.right.right = new TreeNode(3);
root3.left.left.left = new TreeNode(4);
root3.right.right.right = new TreeNode(4);
System.out.println(s.minDepth(root3));
// test4
TreeNode root4 = new TreeNode(1);
root4.left = new TreeNode(2);
root4.left.left = new TreeNode(3);
System.out.println(s.minDepth(root4));
}
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
if (root.left == null && root.right == null) { // left, right 均为 null, 说明 root 是叶子结点
return 1;
}
if (root.left == null) { // left 为 null, 而 right 不为 null, 说明 root 并不是叶子结点,还需要在 right 上查找叶子结点
return minDepth(root.right) + 1;
}
if (root.right == null) {
return minDepth(root.left) + 1;
}
// left, right 均不为 null, 说明还需要在 left 和 right 上查找叶子结点
int leftMinDepth = minDepth(root.left) + 1;
int rightMinDepth = minDepth(root.right) + 1;
return Math.min(leftMinDepth, rightMinDepth);
}
}