【算法學習】二叉樹的最小深度minimum-depth-of-binary-tree

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二叉樹的最小深度牛客網測試

一、 題目描述

求給定二叉樹的最小深度。最小深度是指樹的根結點到最近葉子結點的最短路徑上結點的數量。
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：樹的遍歷，只要能在遍歷過程中記住葉子節點和葉子節點處於的層級（到root有幾個節點即可）
思路2：樹的層序遍歷，層序遍歷過程中遇到葉子節點即可得到結果

三、參考代碼

思路1

public int run(TreeNode root) {
        if (root == null) {
            return 0;
        }
        Map<TreeNode, Integer> map = new HashMap<>();//用於存儲非葉子節點到跟的長度
        Map<TreeNode, Integer> map2 = new HashMap<>();//用於存儲葉子節點到跟的長度
        Stack<TreeNode> l3 = new Stack<>();//用於存儲遍歷中的樹節點
        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;//用於記錄當前層有多少個節點
        Queue<TreeNode> queue = new LinkedTransferQueue<>();//用於存儲遍歷中的樹節點
        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;
    }