[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));
}
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