【算法学习】二叉树的最小深度minimum-depth-of-binary-tree

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二叉树的最小深度牛客网测试

一、 题目描述

求给定二叉树的最小深度。最小深度是指树的根结点到最近叶子结点的最短路径上结点的数量。
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.

二、思路分析

思路1：树的遍历，只要能在遍历过程中记住叶子节点和叶子节点处于的层级（到root有几个节点即可）
思路2：树的层序遍历，层序遍历过程中遇到叶子节点即可得到结果

三、参考代码

思路1

public int run(TreeNode root) {
        if (root == null) {
            return 0;
        }
        Map<TreeNode, Integer> map = new HashMap<>();//用于存储非叶子节点到跟的长度
        Map<TreeNode, Integer> map2 = new HashMap<>();//用于存储叶子节点到跟的长度
        Stack<TreeNode> l3 = new Stack<>();//用于存储遍历中的树节点
        l3.push(root);
        map.put(root, 0);
        while (l3.size() > 0) {
            TreeNode tempRoot = l3.pop();
            TreeNode left = tempRoot.left;
            TreeNode right = tempRoot.right;
            if (left != null) {
                l3.push(left);
                map.put(left, map.get(tempRoot) + 1);
            }
            if (right != null) {
                l3.push(right);
                map.put(right, map.get(tempRoot) + 1);
            }
            if (left == null && right == null) {
                map2.put(tempRoot, map.get(tempRoot) + 1);
            }
        }
        int min = Integer.MAX_VALUE;
        Set<TreeNode> set = map2.keySet();
        for (TreeNode item : set) {
            if (map2.get(item) < min) {
                min = map2.get(item);
            }
        }
        return min;
    }

思路2

public int run(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int layer = 1;//用于记录当前层有多少个节点
        Queue<TreeNode> queue = new LinkedTransferQueue<>();//用于存储遍历中的树节点
        TreeNode layerRightNode = root;
        queue.add(root);
        while (queue.size() > 0) {
            TreeNode tempRoot = queue.remove();
            TreeNode left = tempRoot.left;
            TreeNode right = tempRoot.right;
            if (left != null) {
                queue.add(left);
            }
            if (right != null) {
                queue.add(right);
            }
            if (left == null && right == null) {
                return layer;
            }
            if (layerRightNode == tempRoot) {
                layerRightNode = right == null ? left : right;
                layer++;
            }
        }
        return layer;
    }