代码随想录算法训练营第十七天 | 110、257、404

110. Balanced Binary Tree

题目链接：https://leetcode.cn/problems/balanced-binary-tree/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        if (root == null) return true;
        return getHeght(root) == -1 ? false : true;
    }
    public int getHeght(TreeNode node){
        if (node == null) return 0;
        int l = getHeght(node.left);
        if (l == -1) return -1;
        int r = getHeght(node.right);
        if (r == -1) return -1;
        else return Math.abs(l - r) > 1 ? -1 : 1 + Math.max(l, r);
    }
}
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257. Binary Tree Paths

题目链接：https://leetcode.cn/problems/binary-tree-paths/

方法一 迭代

1 方法思想

2 代码实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    
    public List<String> binaryTreePaths(TreeNode root) {
                List<String> ans = new ArrayList<>();
        if (root == null) return ans;
        Stack<TreeNode> stack = new Stack<>();
        Stack<String> path = new Stack<>();
        stack.push(root);
        path.push(String.valueOf(root.val));

        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            String pathTemp = path.pop();
            if (node.right == null && node.left == null) {
                ans.add(pathTemp);
            }
            if (node.right != null) {
                stack.push(node.right);
                path.add(String.valueOf(new StringBuffer(pathTemp).append("->").append(node.right.val)));
            }
            if (node.left != null) {
                stack.push(node.left);
                path.add(String.valueOf(new StringBuffer(pathTemp).append("->").append(node.left.val)));
            }
        }
        return ans;
    }
}
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3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

方法二 递归

1 方法思想

2 代码实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    
     List<String> ans = new ArrayList<>();

    public List<String> binaryTreePaths(TreeNode root) {

        if (root == null) return ans;
        else {
            if (root.left != null) {
                getPath(root.left, String.valueOf(root.val));
            }
            if (root.right != null) {
                getPath(root.right, String.valueOf(root.val));
            }
            if (root.left == null && root.right == null)ans.add(String.valueOf(root.val));
        }
        return ans;
    }

    public void getPath(TreeNode node, String path) {

        String temPath = path + "->" + String.valueOf(node.val);
        if (node.left != null) {
            getPath(node.left, temPath);
        }
        if (node.right != null) {
            getPath(node.right, temPath);
        }
        if (node.left == null && node.right == null){
            ans.add(temPath);
        }
    }
}
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3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

收获和总结

1.遇到的困难？

2 收获

学习链接

https://programmercarl.com/0704.%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE.html#%E6%80%9D%E8%B7%AF

404. Sum of Left Leaves

题目链接：https://leetcode.cn/problems/sum-of-left-leaves/

方法一 递归

1 方法思想

2 代码实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) return 0;
        int mid = 0;
        int leftTree = sumOfLeftLeaves(root.left);
        int rightTree = sumOfLeftLeaves(root.right);
        if (root.left!=null && root.left.left == null && root.left.right == null){
            mid = root.left.val;
        }
        return leftTree + mid + rightTree;
    }

}
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3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

方法二 迭代

1 方法思想

2 代码实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) return 0;
        int mid = 0;
        int leftTree = sumOfLeftLeaves(root.left);
        int rightTree = sumOfLeftLeaves(root.right);
        if (root.left!=null && root.left.left == null && root.left.right == null){
            mid = root.left.val;
        }
        return leftTree + mid + rightTree;
    }

}
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3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

收获和总结

1.遇到的困难？

2 收获

学习链接

https://programmercarl.com/0704.%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE.html#%E6%80%9D%E8%B7%AF