leetcode 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.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        # traverse tree, if node.left is leaf them sum it
        #if is_leaf(root) and direct is left:
        #    sum it
                
        def sum_left_leaf(node, is_left, sums):
            if not node: return
            if node and (not node.left) and (not node.right) and is_left:
                sums[0] += node.val
                return
            sum_left_leaf(node.left, True, sums)
            sum_left_leaf(node.right, False, sums)
        
        sums = [0]
        sum_left_leaf(root, False, sums)
        return sums[0]                

更简洁的写法：

class Solution(object):
    def sumOfLeftLeaves(self, root):
        if not root:
            return 0
        if root.left and not root.left.left and not root.left.right:
            return root.left.val + self.sumOfLeftLeaves(root.right)
        return self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)

 

迭代解法：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        # traverse tree, if node.left is leaf them sum it
        #if is_leaf(root) and direct is left:
        #    sum it
        ans = 0
        if not root: return ans        
        stack = [(root, False)]
        while stack:
            node, is_left = stack.pop()
            if not node.left and not node.right and is_left:
                ans += node.val
            if node.right: stack.append((node.right, False))
            if node.left: stack.append((node.left, True))
        return ans
                        

另外也可以在迭代过程中绕过叶子结点：

Iterative method. Here for each node in the tree we check whether its left child is a leaf. If it is true, we add its value to answer, otherwise add left child to the stack to process it later. For right child we add it to stack only if it is not a leaf.

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) {
            if (node.left.left == null && node.left.right == null)
                ans += node.left.val;
            else
                stack.push(node.left);
        }
        if(node.right != null) {
            if (node.right.left != null || node.right.right != null)
                stack.push(node.right);
        }
    }
    return ans;
}