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112. Path Sum

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	`1`	`+/**`
	`2`	`+ * Definition for a binary tree node.`
	`3`	`+ * public class TreeNode {`
	`4`	`+ * int val;`
	`5`	`+ * TreeNode left;`
	`6`	`+ * TreeNode right;`
	`7`	`+ * TreeNode() {}`
	`8`	`+ * TreeNode(int val) { this.val = val; }`
	`9`	`+ * TreeNode(int val, TreeNode left, TreeNode right) {`
	`10`	`+ * this.val = val;`
	`11`	`+ * this.left = left;`
	`12`	`+ * this.right = right;`
	`13`	`+ * }`
	`14`	`+ * }`
	`15`	`+ */`
	`16`	`+classSolution {`
	`17`	`+privatebooleanisLeafNode(TreeNodenode) {`
	`18`	`+return ((node.left ==null) && (node.right ==null));`
	`19`	`+ }`
	`20`	`+`
	`21`	`+publicbooleanhasPathSum(TreeNoderoot,inttargetSum) {`
	`22`	`+// Edge case: No nodes`
	`23`	`+if(root ==null) {`
	`24`	`+returnfalse;`
	`25`	`+ }`
	`26`	`+`
	`27`	`+targetSum -=root.val;`
	`28`	`+if(isLeafNode(root)) {`
	`29`	`+return (targetSum ==0);`
	`30`	`+ }`
	`31`	`+returnhasPathSum(root.left,targetSum) \|\|hasPathSum(root.right,targetSum);`
	`32`	`+ }`
	`33`	`+}`

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