Leetcode 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.

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解题思路：

方法一： 递归解法，注意如果有个节点只有一边孩子时，不能返回0，要返回另外一半边的depth。 

方法二： BFS, 求layer 的数目，当一个节点没有孩子的时候，返回depth。

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Java code

方法一： 递归解法

 public int minDepth(TreeNode root) {
        if(root == null) {
            return 0;
        }
        if(root.left == null && root.right == null) {
            return 1;
        }
        int leftmin = minDepth(root.left);
        int rightmin = minDepth(root.right);
        if(root.right == null) {
            return leftmin + 1;
        }
        if(root.left == null) {
            return rightmin + 1;
        }else {
            return Math.min(leftmin, rightmin) + 1;
        }
    }

或者

public int minDepth(TreeNode root) {
        if(root == null)
            return 0;
        int minleft = minDepth(root.left);
        int minright = minDepth(root.right);
        if(minleft==0 || minright==0)
            return minleft>=minright?minleft+1:minright+1;
        return Math.min(minleft,minright)+1;
    }

方法二： BFS, 求layer 的数目，当一个节点没有孩子的时候，返回depth。

public int minDepth(TreeNode root) {
        //use BFS
        if(root == null){
            return 0;
        }
        int depth = 1;
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int curNum = 1; //num of node left in current level
        int nextNum = 0; // num of nodes in next level
        while(!queue.isEmpty()){
            TreeNode cur = queue.poll();
            curNum--;
            
            if(cur.left == null && cur.right == null) {
                return depth;
            }
            if(cur.left != null) {
                queue.add(cur.left);
                nextNum++;
            }
            if(cur.right != null){
                queue.add(cur.right);
                nextNum++;
            }
            if(curNum == 0) {
                curNum = nextNum;
                nextNum = 0;
                depth++;
            }
        }
        return depth;    
    }

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20160601

recursion

public class Solution {
    public int minDepth(TreeNode root) {
        //use recursion every time get min
        //base case
        if (root == null) {
            return 0;
        } else if (root.left == null && root.right == null) {
            return 1;
        }
        int num = 0;
        
        if (root.left != null && root.right != null) {
            num = Math.min(minDepth(root.left), minDepth(root.right)) + 1;
        } else if (root.left == null && root.right != null) {
            num = minDepth(root.right) + 1;
        } else if (root.left != null && root.right == null) {
            num = minDepth(root.left) + 1;
        }
        return num;
    }
}

Reference:

1. http://www.cnblogs.com/springfor/p/3879680.html