404. Sum of Left Leaves

404. Sum of Left Leaves

Find the sum of all left leaves in a given binary tree.

Example:

    3
   / \
  9  20
    /  \
   15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

My Solutions:

左叶节点的定义是本身是叶(不是root，且左右孩子为null），且是左节点。因此，只有满足这两个条件的节点值才会返回。

方法1：dfs recursive

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

方法2：dfs iterative。dfs一律用stack做，因为最后进的最先出，所以只要node.left != null，就一直压栈

public class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        
        if (root == null) return 0;
        int ans = 0;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
    
        while (!stack.empty()) {
            TreeNode node = stack.pop();
            if (node.left != null) {
                //这个节点就是左叶节点
                if (node.left.left == null && node.left.right == null) 
                    ans += node.left.val;
                else
                    stack.push(node.left);
            }
            // 只有有右节点不为null且有孩子时，才压这个节点。如果本身是个右叶节点，不需要压
            if (node.right != null) { 
                if (node.right.left != null || node.right.right != null)
                    stack.push(node.right);
            }
        }
        return ans;
    }
}

方法3：bfs iterative。bfs一律用queue做，因为先进先出，所以总是loop完一层，再进入下一层

public class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) return 0;
        int res = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while(!queue.isEmpty()) {
            TreeNode curr = queue.poll();
            // curr.left is a left leaf
            if (curr.left != null && curr.left.left == null && curr.left.right == null) res += curr.left.val;
            if (curr.left != null) queue.offer(curr.left);
            if (curr.right != null) queue.offer(curr.right);
        }
        return res;
    }
}