[LeetCode]104. MaximumDepthOfBinaryTree&110. BalancedBinaryTree&111. MinimumDepthOfBinaryTree

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本文介绍了二叉树的最大深度、最小深度及平衡性的算法实现，包括递归和迭代两种方式，并通过具体示例说明了如何求解二叉树的不同深度。

104 . Maximum Depth of Binary Tree
Easy

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.

2ms:

public int maxDepth(TreeNode root) {
    return getDep(root,0);
}
private int getDep(TreeNode p,int now){
    if(p!=null){
        int a = getDep(p.left,(now+1));
        int b = getDep(p.right,(now+1));
        return Math.max(a,b);
    }
    return now;
}

1ms:

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

110 . Balanced Binary Tree
Easy

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.

3ms:

 public boolean isBalanced(TreeNode root) {
        if(root==null)
            return true;

        if(Math.abs(maxDepth(root.left)-maxDepth(root.right))>1)
            return false;

        return isBalanced(root.left)&&isBalanced(root.right);

    }

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

111 . Minimum Depth of Binary Tree
Easy

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],

/ \
9 20
/ \
15 7
return its minimum depth = 2.

DFS

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

BFS

public int minDepth(TreeNode root) {
        if(root == null) return 0;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        int depth = 1;
        while(!queue.isEmpty()) {
            int size = queue.size();
            while(size-- > 0) {
                TreeNode p = queue.poll();
                if (p.left == null && p.right == null) {
                    return depth;
                }
                if (p.left != null) {
                    queue.offer(p.left);
                }
                if (p.right != null) {
                    queue.offer(p.right);
                }
            }

            depth++;
        }
        return depth;
    }