[Leetcode][第404题][JAVA][左叶子之和][DFS][BFS]

【问题描述】[简单]

在这里插入图片描述

【解答思路】

1. DFS

递进思想一步一步递进

/先序遍历求所有节点值之和

public int sumOfTrees(TreeNode root) {if (root == null) {return 0;}int leave = root.val;int left = sumOfTrees(root.left);int right = sumOfTrees(root.right);return left + right + leave;}

在求节点之和的基础上，我们再来求所有叶子节点之和。

// 先序遍历求所有叶子节点值之和public int sumOfLeaves(TreeNode root) {if (root == null) {return 0;}int leave = 0;// 叶子节点if (root.left == null && root.right == null) {leave = root.val;}int left = sumOfLeaves(root.left);int right = sumOfLeaves(root.right);return left + right + leave;}

在求叶子节点之和的基础上，我们再来求所有左叶子节点之和。

当前节点是不是左节点，可以通过父节点标识。

    public int sumOfLeftLeaves(TreeNode root) {return sumOfLeftLeavesHelper(root, false);}// 先序遍历求所有左叶子节点值之和public int sumOfLeftLeavesHelper(TreeNode root, boolean flag) {if (root == null) {return 0;}int leave = 0;// 左叶子节点if (flag && root.left == null && root.right == null) {leave = root.val;}int left = sumOfLeftLeavesHelper(root.left, true);int right = sumOfLeftLeavesHelper(root.right, false);return left + right + leave;}

通俗易懂版本

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {int sum=0;public int sumOfLeftLeaves(TreeNode root) {if(root==null) return 0;dfs(root,0);return sum;}public void dfs(TreeNode root, int fPath){if(root.left==null && root.right==null && fPath==1){sum+=root.val;}if(root.right!=null){dfs(root.right,-1);}if(root.left!=null){dfs(root.left,1);}}}

2. BFS

时间复杂度：O(N) 空间复杂度：O(1)

class Solution {public int sumOfLeftLeaves(TreeNode root) {if (root == null) {return 0;}Queue<TreeNode> queue = new LinkedList<TreeNode>();queue.offer(root);int ans = 0;while (!queue.isEmpty()) {TreeNode node = queue.poll();if (node.left != null) {if (isLeafNode(node.left)) {ans += node.left.val;} else {queue.offer(node.left);}}if (node.right != null) {if (!isLeafNode(node.right)) {queue.offer(node.right);}}}return ans;}public boolean isLeafNode(TreeNode node) {return node.left == null && node.right == null;}
}