day18 - 二叉树 part05

513. 找树左下角的值

详解

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int max_depth = INT_MIN;
    void search(TreeNode* node, int depth, int& result){
        //前序遍历 中左右
        //父节点控制结束
        if(node->left == NULL && node->right == NULL){
            if(depth > max_depth){
                max_depth = depth;
                result = node->val;
            }
            return;
        }
        if(node->left){
            depth++;
            search(node->left, depth, result);
            depth--;//深度回溯
        }
        if(node->right){
            depth++;
            search(node->right, depth, result);
            depth--;//深度回溯
        }
        return;
    }

    int findBottomLeftValue_recursion(TreeNode* root) {
        int result;
        search(root, 0, result);
        return result;
    }

    int findBottomLeftValue(TreeNode* root) {
        int result;
        queue<TreeNode*> queue_1;
        if(!root) return 0;
        queue_1.push(root);
        while(!queue_1.empty()){
            int size = queue_1.size();//cout<<size<<endl;
            for(int i=0; i< size; i++){
                TreeNode* tmp = queue_1.front();
                if(i == 0)
                    result = tmp->val;
                queue_1.pop();
                if(tmp->left) queue_1.push(tmp->left);
                if(tmp->right) queue_1.push(tmp->right);
            }
            
        }
        return result;
    }
};

 

112. 路径总和

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool search(TreeNode* node, int total, int target){
        if(node->left == NULL && node->right == NULL){
            if(total == target)
                return true;
            else
                return false;
        }
        bool left_ok = false;
        bool right_ok = false;
        if(node->left){
            total += node->left->val;
            left_ok = search(node->left, total, target);
            total -= node->left->val;
        }
        if(node->right){
            total += node->right->val;
            right_ok = search(node->right, total, target);
            total -= node->right->val;
        }
        return left_ok || right_ok;
    }

    bool hasPathSum(TreeNode* root, int targetSum) {
        if(!root) return false;
        return search(root, root->val, targetSum);
    }
};

113. 路径总和 II

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void search(TreeNode* node, vector<int>& path, vector<vector<int>>& result, int total, int target){
        if(node->left == NULL && node->right == NULL){
            if(total == target){
                result.push_back(path);
            }
            return;
        }
        if(node->left){
            total += node->left->val;
            path.push_back(node->left->val);
            search(node->left, path, result, total, target);
            total -= node->left->val;
            path.pop_back();
        }
        if(node->right){
            total += node->right->val;
            path.push_back(node->right->val);
            search(node->right, path, result, total, target);
            total -= node->right->val;
            path.pop_back();
        }
        return;
    }

    vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
        vector<int> path;
        vector<vector<int>> result;
        if(root){
            path.push_back(root->val);
            search(root, path, result, root->val, targetSum);
        } 
        return result;
    }
};

106. 从中序与后序遍历序列构造二叉树

class Solution {
private:
    TreeNode* traversal (vector<int>& inorder, vector<int>& postorder) {
        if (postorder.size() == 0) return NULL;

        // 后序遍历数组最后一个元素，就是当前的中间节点
        int rootValue = postorder[postorder.size() - 1];
        TreeNode* root = new TreeNode(rootValue);

        // 叶子节点
        if (postorder.size() == 1) return root;

        // 找到中序遍历的切割点
        int delimiterIndex;
        for (delimiterIndex = 0; delimiterIndex < inorder.size(); delimiterIndex++) {
            if (inorder[delimiterIndex] == rootValue) break;
        }

        // 切割中序数组
        // 左闭右开区间：[0, delimiterIndex)
        vector<int> leftInorder(inorder.begin(), inorder.begin() + delimiterIndex);
        // [delimiterIndex + 1, end)
        vector<int> rightInorder(inorder.begin() + delimiterIndex + 1, inorder.end() );

        // postorder 舍弃末尾元素
        postorder.resize(postorder.size() - 1);

        // 切割后序数组
        // 依然左闭右开，注意这里使用了左中序数组大小作为切割点
        // [0, leftInorder.size)
        vector<int> leftPostorder(postorder.begin(), postorder.begin() + leftInorder.size());
        // [leftInorder.size(), end)
        vector<int> rightPostorder(postorder.begin() + leftInorder.size(), postorder.end());

        root->left = traversal(leftInorder, leftPostorder);
        root->right = traversal(rightInorder, rightPostorder);

        return root;
    }
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if (inorder.size() == 0 || postorder.size() == 0) return NULL;
        return traversal(inorder, postorder);
    }
};