首页 > 代码库 > 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 7There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
Recusive:
public int SumOfLeftLeaves(TreeNode root) { return DFS(root,false); } public int DFS(TreeNode root, bool left) { if(root == null) return 0; if(root.left==null && root.right==null ) return left?root.val:0; return DFS(root.left,true)+DFS(root.right,false); }
Iteration:
public int SumOfLeftLeaves(TreeNode root) { if(root == null) return 0; var stack = new Stack<TreeNode>(); int sum =0; stack.Push(root); while(stack.Count()>0) { var a = stack.Pop(); if(a.left != null) { if(a.left.left ==null && a.left.right==null) sum += a.left.val; else stack.Push(a.left); } if(a.right != null) stack.Push(a.right); } return sum; }
404. Sum of Left Leaves
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。