[Leetcode][JAVA] Minimum Depth of Binary Tree && Balanced Binary Tree && Maximum Depth of Binary Tree

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.

比较左子树和右子树的深度，取小的那个返回上来，并+1。

需要注意的是如果没有左子树或右子树。那么无条件取存在的那一边子树深度，并+1.

如果左子树和右子树都没有，那么就是叶子节点，返回深度1.

如果root自身为null,返回0

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

Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

新定义一个函数，表示树的高度，如果是平衡树的话。不是平衡树则置高度为-1. 运用分治法递归，最后得到结果为-1则不是平衡的。基本情况，root==null则高度为0

     public boolean isBalanced(TreeNode root) {
         if(height(root)==-1)
             return false;
         return true;
     }
     public int height(TreeNode root)
     {
         if(root==null)
             return 0;
         int left = height(root.left);
         int right = height(root.right);
         if(left==-1 || right==-1)
             return -1;
         if(Math.abs(left-right)>1)
             return -1;
         return Math.max(left,right)+1;
     }

Maximum Depth of Binary Tree

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

求最大深度方法也是一样,更简单了：

     public int maxDepth(TreeNode root) {
         if(root==null)
             return 0;
         int left = maxDepth(root.left);
         int right = maxDepth(root.right);
         return (left>right?(left+1):(right+1));
     }