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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.
求一棵树的最短路径,同样用递归来解决。注意的是如果遇到一个节点只有一个孩子,这时我们要继续往下搜,要不然不符合题目的规定,同样的还有处理最大路径的问题,都是采用递归解决。代码如下:
最短路径:
最长路径:
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
求一棵树的最短路径,同样用递归来解决。注意的是如果遇到一个节点只有一个孩子,这时我们要继续往下搜,要不然不符合题目的规定,同样的还有处理最大路径的问题,都是采用递归解决。代码如下:
最短路径:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public int minDepth(TreeNode root) { if(root == null) return 0; return minDepth(root, 1); } public int minDepth(TreeNode root, int depth) { if(root == null) return 0; if(root.left != null && root.right != null) return Math.min(minDepth(root.left, depth + 1), minDepth(root.right, depth + 1)); if(root.left != null && root.right == null) return minDepth(root.left, depth + 1); if(root.left == null && root.right != null) return minDepth(root.right, depth + 1); return depth; } }
最长路径:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public int maxDepth(TreeNode root) { if(root == null) return 0; return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1; } }
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