左叶子节点之和

Author: llb19911212

Tree/llb/404. Sum of Left Leaves.cpp

@@ -0,0 +1,94 @@
+/*
+问题描述：
+
+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.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    
+    //思路一：递归
+    /*
+    int sumOfLeftLeaves(TreeNode* root) {
+        if (!root || !root->left && !root->right) 
+            return 0;
+        
+        int res = 0;   
+        //分别左右两支开始进行
+        help(root->left, true, res);
+        help(root->right, false, res);
+        
+        return res;
+    }
+    
+    void help(TreeNode* root, bool isLeft, int &res) {
+        if (!root)
+            return;
+        if (!root->left && !root->right && isLeft)   //如果是叶子节点并且是左孩子
+            res += root->val;
+        help(root->left, true, res);
+        help(root->right, false, res);
+    }
+    */
+    
+    //写法二：
+    /*
+    int sumOfLeftLeaves(TreeNode* root) {
+        if (!root) 
+            return 0;
+        
+        //如果该节点的左孩子存在，但它是叶子节点。那就加上自己再计算右子树
+        if (root->left && !root->left->left && !root->left->right)
+            return root->left->val + sumOfLeftLeaves(root->right);
+        
+        return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
+    }
+    */
+    
+    //思路二：利用层序遍历，逐个相加计算
+    int sumOfLeftLeaves(TreeNode* root) {
+        if (!root || !root->left && !root->right) {
+            return 0;
+        }
+        
+        int res = 0;
+        queue<TreeNode*> q;
+        q.push(root);
+        while(!q.empty()) {
+            TreeNode* node = q.front();
+            q.pop();
+            //如果是左边叶子节点
+            if (node->left && !node->left->left && !node->left->right)
+                res += node->left->val;
+            if (node->left)
+                q.push(node->left);
+            if (node->right)
+                q.push(node->right);
+        }
+        
+        return res;
+    }
+    
+    
+};