111. Minimum Depth of Binary Tree.md

111. Minimum Depth of Binary Tree

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.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its minimum depth = 2.

Solution

有个trick是不能直接用height = 1+min(self.minDepth(root.left),self.minDepth(root.right))。因为这样的话不能保证深度是从叶结点开始算的。比如下面的情况就会返回1而不是2.

  3
 /
1

所以需要讨论左右是不是None的情况。

Code

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def minDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root is None:
            return 0
        
        if root.left and root.right:
            height = 1+min(self.minDepth(root.left),self.minDepth(root.right))
        elif root.left:
            height = 1+self.minDepth(root.left)
        elif root.right:
            height = 1+self.minDepth(root.right)
        else:
            height = 1    
        return height

Code Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int res = Integer.MAX_VALUE;
    private void helper(TreeNode root, int cur)
    {
        if(root == null)
            return;
        else if(root.left == null && root.right == null)
            res = (cur+1)<res? cur+1:res;
        else
        {
            helper(root.right,cur+1);
            helper(root.left,cur+1);   
        }
        
    }
    public int minDepth(TreeNode root) {
        int cur = 0;
        if(root==null)
            return 0;
            
        helper(root,cur);
        return res;
    }
}