博文中会简要介绍Leetcode P0111题目分析及解题思路。
“Minimum Depth of Binary Tree”是一道比较基础的题目，利用深度优先搜索，这里采取递归回溯的方法，每次获取当前树结点的左右子树高度，然后取其中的最小值。需要注意的是，若是非叶子结点的树结点，要注意若其左或者有子树不存在，则不存在的子树高度要设置为正无穷，以保证所有高度都是根结点到叶子结点的高度
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.
Note: A leaf is a node with no children.
以下是Java的题解代码实现。
/**
* 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;
return this.traverse(root);
}
private int traverse(TreeNode curr) {
if (curr.left == null && curr.right == null)
return 1;
int leftHeight = Integer.MAX_VALUE, rightHeight = Integer.MAX_VALUE;
if (curr.left != null)
leftHeight = this.traverse(curr.left);
if (curr.right != null)
rightHeight = this.traverse(curr.right);
return Math.min(leftHeight, rightHeight)+1;
}
}
以下是C++的题解代码实现。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode* root) {
if (!root)
return 0;
if (!root->left && !root->right)
return 1;
int lh = INT_MAX, rh = INT_MAX;
if (root->left)
lh = this->minDepth(root->left);
if (root->right)
rh = this->minDepth(root->right);
return min(lh, rh)+1;
}
};