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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.
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; }
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原文地址：http://www.cnblogs.com/renyualbert/p/5911358.html
