创作目的：为了方便自己后续复习重点，以及养成写博客的习惯。

一、找树左下角的值

思路：采用递归

ledcode题目：https://leetcode.cn/problems/find-bottom-left-tree-value/description/

AC代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 void dfs(const struct TreeNode *root ,int height, int *curVal ,int *curHeight) {
     if(root == NULL) {
         return ;
     }
     height++;
     dfs(root->left,height,curVal,curHeight);
     dfs(root->right,height,curVal,curHeight);
     if(height > *curHeight) {
         *curHeight = height;
         *curVal    = root->val;
     }
 }
int findBottomLeftValue(struct TreeNode* root) {
    int curVal,curHeight = 0;
    dfs(root,0,&curVal,&curHeight);
    return curVal;
}

二、路径总和

思路：采用递归，判断targetSum == root->val，sum的值每次更新减去前一个val即可。

lecode题目：https://leetcode.cn/problems/path-sum/

AC代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool hasPathSum(struct TreeNode* root, int targetSum) {
    if(root == NULL) {
        return false;
    }
    if(root->left == NULL && root->right == NULL) {
        return targetSum == root->val;
    }
    return hasPathSum(root->left,targetSum - root->val) || hasPathSum(root->right,targetSum - root->val);
}

三、从中序与后序遍历序列构造二叉树

思路：采用迭代的方法，代码参考的是ledcode题解

ledcode题目：https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/

AC代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 struct TreeNode* createTreeNode(int val) {
     struct TreeNode* ret = malloc(sizeof(struct TreeNode));
     ret->val = val;
     ret->left = ret->right = NULL;
     return ret;
 }
struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize) {
    if(postorderSize == 0) {
        return NULL;
    }
    struct TreeNode* root = createTreeNode(postorder[postorderSize - 1]);
    struct TreeNode** s = malloc(sizeof(struct TreeNode*)* 10001);
    int top = 0;
    s[top++] = root;
    int inorderIndex = inorderSize - 1;
    for(int i = postorderSize - 2;i >= 0;i--) {
        int postorderVal = postorder[i];
        struct TreeNode* node = s[top - 1];
        if(node->val != inorder[inorderIndex]) {
            node->right = createTreeNode(postorderVal);
            s[top++] = node->right;
        }else {
            while(top > 0 && s[top - 1]->val == inorder[inorderIndex]) {
                node = s[--top];
                inorderIndex--;
            }
            node->left = createTreeNode(postorderVal);
            s[top++] = node->left;
        }
    }
    return root;
}

全篇后记：

虽然难度在不断增加，但坚持下去总会有收获。