110. Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.
A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

题目解析：

• 可以用递归求左右子树的高度差
• 使用-1表示该子树不平衡

代码：

class Solution {
    public boolean isBalanced(TreeNode root) {
        return depth(root) == -1 ? false : true;
    }

    public int depth(TreeNode root) {
        if (root == null) return 0;
        int left = depth(root.left);
        int right = depth(root.right);
        if (left == -1 || right == -1 || Math.abs(left - right) > 1) {
            return -1;
        }
        return Math.max(left, right) + 1;
    }
}

257. Binary Tree Paths

Given the root of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.

题目解析：

• 求二叉树所有路径使用DFS比较合适

代码：

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<>();
        dfs(root, new ArrayList<>(), result);
        return result;
    }

    public void dfs(TreeNode root, List<String> path, List<String> result) {
        path.add(String.valueOf(root.val));
        if (root.left == null && root.right == null) {
            String s = path.stream().collect(Collectors.joining("->"));
            result.add(s);
        }
        if (root.left != null) {
            dfs(root.left, path, result);
        }
        if (root.right != null) {
            dfs(root.right, path, result);
        }
        path.remove(path.size() - 1);
    }
}

404. Sum of Left Leaves

Given the root of a binary tree, return the sum of all left leaves.

A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.

题目解析：

• 节点是否为叶子节点并且为父节点的左节点， 可使用flag标记是否为左节点

代码：

class Solution {

    public int sumOfLeftLeaves(TreeNode root) {
        return dfs(root, false);
    }
    public int dfs(TreeNode root, boolean isLeft) {
        if (root == null) {
            return 0;
        }
        int sum = 0;
        if (root.left != null) {
            sum += dfs(root.left, true);
        }
        if (root.right != null) {
            sum += dfs(root.right, false);
        }
        if (isLeft && root.left == null && root.right == null) {
            sum += root.val;
        }
        return sum;
    }
}

class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int sum = 0;
        if (root.left != null) {
            if (isLeaf(root.left)) {
                sum += root.left.val;
            } else {
                sum += sumOfLeftLeaves(root.left);
            }
        }
        if (root.right != null) {
            sum += sumOfLeftLeaves(root.right);
        }
        return sum;
    }

    public boolean isLeaf(TreeNode root) {
        if (root.left == null && root.right == null) {
            return true;
        }
        return false;
    }
}