From 61f88b332c31687361cdb4aa09f83c1da874d4b5 Mon Sep 17 00:00:00 2001 From: youngyangyang04 <826123027@qq.com> Date: Wed, 7 Oct 2020 18:17:46 +0800 Subject: [PATCH] Update --- README.md | 3 + pics/513.找树左下角的值.png | Bin 0 -> 13637 bytes pics/513.找树左下角的值1.png | Bin 0 -> 19727 bytes problems/0513.找树左下角的值.md | 162 ++++++++++++++++++++++++- 4 files changed, 159 insertions(+), 6 deletions(-) create mode 100644 pics/513.找树左下角的值.png create mode 100644 pics/513.找树左下角的值1.png diff --git a/README.md b/README.md index 18d36d58..5d91b0d2 100644 --- a/README.md +++ b/README.md @@ -95,6 +95,9 @@ * [二叉树:我平衡么?](https://mp.weixin.qq.com/s/isUS-0HDYknmC0Rr4R8mww) * [二叉树:找我的所有路径?](https://mp.weixin.qq.com/s/Osw4LQD2xVUnCJ-9jrYxJA) * [还在玩耍的你,该总结啦!(本周小结之二叉树)](https://mp.weixin.qq.com/s/QMBUTYnoaNfsVHlUADEzKg) + * [二叉树:以为使用了递归,其实还隐藏着回溯](https://mp.weixin.qq.com/s/ivLkHzWdhjQQD1rQWe6zWA) + * [二叉树:做了这么多题目了,我的左叶子之和是多少?](https://mp.weixin.qq.com/s/gBAgmmFielojU5Wx3wqFTA) + (持续更新中....) diff --git 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HcmV?d00001 diff --git a/problems/0513.找树左下角的值.md b/problems/0513.找树左下角的值.md index 1da53893..ec4d041c 100644 --- a/problems/0513.找树左下角的值.md +++ b/problems/0513.找树左下角的值.md @@ -1,5 +1,104 @@ -## C++代码递归 +> 我的左下角的数值是多少? + +# 513.找树左下角的值 + +给定一个二叉树,在树的最后一行找到最左边的值。 + +示例 1: + + + +示例 2: + + + +# 思路 + +本地要找出树的最后一行找到最左边的值。此时大家应该想起用层序遍历是非常简单的了,反而用递归的话会比较难一点。 + +我们依然还是先介绍递归法。 + +## 递归 + +咋眼一看,这道题目用递归的话就就一直向左遍历,最后一个就是答案呗? + +没有这么简单,一直向左遍历到最后一个,它未必是最后一行啊。 + +我们来分析一下题目:在树的**最后一行**找到**最左边的值**。 + +首先要是最后一行,然后是最左边的值。 + +如果使用递归法,如何判断是最后一行呢,其实就是深度最大的叶子节点一定是最后一行。 + +如果对二叉树深度和高度还有点疑惑的话,请看:[二叉树:我平衡么?](https://mp.weixin.qq.com/s/isUS-0HDYknmC0Rr4R8mww)。 + +所以要找深度最大的叶子节点。 + +那么如果找最左边的呢?可以使用前序遍历,这样才先优先左边搜索,然后记录深度最大的叶子节点,此时就是树的最后一行最左边的值。 + +递归三部曲: + +1. 确定递归函数的参数和返回值 + +参数必须有要遍历的树的根节点,还有就是一个int型的变量用来记录最长深度。 这里就不需要返回值了,所以递归函数的返回类型为void。 + +本题还需要类里的两个全局变量,maxLen用来记录最大深度,maxleftValue记录最大深度最左节点的数值。 + +代码如下: + +``` +int maxLen = INT_MIN; // 全局变量 记录最大深度 +int maxleftValue; // 全局变量 最大深度最左节点的数值 +void traversal(TreeNode* root, int leftLen) +``` + +有的同学可能疑惑,为啥不能递归函数的返回值返回最长深度呢? + +其实很多同学都对递归函数什么时候要有返回值,什么时候不能有返回值很迷茫。 + +**如果需要遍历整颗树,递归函数就不能有返回值。如果需要遍历某一条固定路线,递归函数就一定要有返回值!** + +初学者可能对这个结论不太理解,别急,后面我会安排一道题目专门讲递归函数的返回值问题。这里大家暂时先了解一下。 + +本题我们是要遍历整个树找到最深的叶子节点,需要遍历整颗树,所以递归函数没有返回值。 + +2. 确定终止条件 + +当遇到叶子节点的时候,就需要统计一下最大的深度了,所以需要遇到叶子节点来更新最大深度。 + +代码如下: + +``` +if (root->left == NULL && root->right == NULL) { + if (leftLen > maxLen) { + maxLen = leftLen; // 更新最大深度 + maxleftValue = root->val; // 最大深度最左面的数值 + } + return; +} +``` + +3. 确定单层递归的逻辑 + +在找最大深度的时候,递归的过程中依然要使用回溯,代码如下: + +``` + // 中 +if (root->left) { // 左 + leftLen++; // 深度加一 + traversal(root->left, leftLen); + leftLen--; // 回溯,深度减一 +} +if (root->right) { // 右 + leftLen++; // 深度加一 + traversal(root->right, leftLen); + leftLen--; // 回溯,深度减一 +} +return; +``` + +完整代码如下: ``` class Solution { @@ -9,20 +108,20 @@ public: void traversal(TreeNode* root, int leftLen) { if (root->left == NULL && root->right == NULL) { if (leftLen > maxLen) { - maxleftValue = root->val; maxLen = leftLen; + maxleftValue = root->val; } return; } if (root->left) { leftLen++; traversal(root->left, leftLen); - leftLen--; + leftLen--; // 回溯 } if (root->right) { leftLen++; traversal(root->right, leftLen); - leftLen--; + leftLen--; // 回溯 } return; } @@ -33,7 +132,48 @@ public: }; ``` -## C++代码层序遍历 +当然回溯的地方可以精简,精简代码如下: + +``` +class Solution { +public: + int maxLen = INT_MIN; + int maxleftValue; + void traversal(TreeNode* root, int leftLen) { + if (root->left == NULL && root->right == NULL) { + if (leftLen > maxLen) { + maxLen = leftLen; + maxleftValue = root->val; + } + return; + } + if (root->left) { + traversal(root->left, leftLen + 1); // 隐藏着回溯 + } + if (root->right) { + traversal(root->right, leftLen + 1); // 隐藏着回溯 + } + return; + } + int findBottomLeftValue(TreeNode* root) { + traversal(root, 0); + return maxleftValue; + } +}; +``` + +如果对回溯部分精简的代码 不理解的话,可以看这篇[二叉树:找我的所有路径?](https://mp.weixin.qq.com/s/Osw4LQD2xVUnCJ-9jrYxJA)和[二叉树:以为使用了递归,其实还隐藏着回溯](https://mp.weixin.qq.com/s/ivLkHzWdhjQQD1rQWe6zWA) 。这两篇文章详细分析了回溯隐藏在了哪里。 + + +## 迭代法 + +本题使用层序遍历再合适不过了,比递归要好理解的多! + +只需要记录最后一行第一个节点的数值就可以了。 + +如果对层序遍历不了解,看这篇[二叉树:层序遍历登场!](https://mp.weixin.qq.com/s/Gb3BjakIKGNpup2jYtTzog),这篇里也给出了层序遍历的模板,稍作修改就一过刷了这道题了。 + +代码如下: ``` class Solution { @@ -47,7 +187,7 @@ public: for (int i = 0; i < size; i++) { TreeNode* node = que.front(); que.pop(); - if (i == 0) result = node->val; + if (i == 0) result = node->val; // 记录最后一行第一个元素 if (node->left) que.push(node->left); if (node->right) que.push(node->right); } @@ -56,3 +196,13 @@ public: } }; ``` + +# 总结 + +本题涉及如下几点: + +* 递归求深度的写法,我们在[二叉树:我平衡么?](https://mp.weixin.qq.com/s/isUS-0HDYknmC0Rr4R8mww)中详细的分析了深度应该怎么求,高度应该怎么求。 +* 递归中其实隐藏了回溯,在[二叉树:以为使用了递归,其实还隐藏着回溯](https://mp.weixin.qq.com/s/ivLkHzWdhjQQD1rQWe6zWA)中讲解了究竟哪里使用了回溯,哪里隐藏了回溯。 +* 层次遍历,在[二叉树:层序遍历登场!](https://mp.weixin.qq.com/s/Gb3BjakIKGNpup2jYtTzog)深度讲解了二叉树层次遍历。 + +所以本题涉及到的点,我们之前都讲解过,这些知识点需要同学们灵活运用,这样就举一反三了。